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Integration

1.10 Partial Fractions

29 problems · hints, answers and solutions shown beside each one

Recall that we are using log⁡x\log x to denote the logarithm of xx with base ee. In other courses it is often denoted ln⁡x\ln x.

1Stage 1Conceptual

Do you understand the idea? Usually little or no calculation.

Q1Stage 1

Below are the graphs of four different quadratic functions. For each quadratic function, decide whether it is: (i) irreducible, (ii) the product of two distinct linear factors, or (iii) the product of a repeated linear factor (and possibly a constant).

Figure from prob_s1.10, line 4

Figure from prob_s1.10, line 4

Figure from prob_s1.10, line 9

Figure from prob_s1.10, line 9

Figure from prob_s1.10, line 14

Figure from prob_s1.10, line 14

Figure from prob_s1.10, line 19

Figure from prob_s1.10, line 19

Hint

If a quadratic function can be factored as (ax+b)(cx+d)(ax+b)(cx+d) for some constants a,b,c,da,b,c,d, then it has roots −ba-\frac{b}{a} and −dc-\frac{d}{c}.

Answer

(a) (iii) (b) (ii) (c) (ii) (d) (i)

Full solution

If a quadratic function can be factored as (ax+b)(cx+d)(ax+b)(cx+d) for some constants a,b,c,da,b,c,d, then it has roots −ba-\frac{b}{a} and −dc-\frac{d}{c}. So, if a quadratic function has no roots, it is irreducible: this is the case for the function in graph (d).

If a quadratic function has two different roots, then (ax+b)≠α(cx+d)(ax+b) \neq \alpha(cx+d) for any constant α\alpha. That is, the quadratic function is the product of distinct linear factors. This is the case for the functions graphed in (b) and (c), since these each have two distinct places where they cross the xx-axis.

Finally, if a quadratic function has precisely one root, then ba=dc\frac{b}{a}=\frac{d}{c}, so:

(ax+b)(cx+d)=a(x+ba)(cx+d)=a(x+dc)(cx+d)=ac(cx+d)(cx+d)\begin{align*} (ax+b)(cx+d)&=a(x+\tfrac{b}{a})(cx+d) = a(x+\tfrac{d}{c})(cx+d) = \tfrac{a}{c}(cx+d)(cx+d) \end{align*}

That is, the quadratic function is the product of a repeated linear factor, and a constant ac\frac{a}{c} (which might simply be ac=1\frac{a}{c}=1).

Q2Stage 1Past exam · 2016Q4

Write out the general form of the partial-fractions decomposition of x3+3(x2−1)2(x2+1)\displaystyle\frac{x^3+3}{(x^2-1)^2(x^2+1)}. You need not determine the values of any of the coefficients.

Hint

Review Equations 1.10.7 through 1.10.11 of the CLP-2 text. Be careful to fully factor the denominator.

Answer

Ax−1+B(x−1)2+Cx+1+D(x+1)2+Ex+Fx2+1\displaystyle\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{Ex+F}{x^2+1}

Full solution

Our first step is to fully factor the denominator:

(x2−1)2(x2+1)=(x−1)2(x+1)2(x2+1)(x^2-1)^2(x^2+1) = (x-1)^2(x+1)^2(x^2+1)

Once a term is linear, it can't be factored further; for quadratic terms, we should check that they are irreducible. Since x2+1x^2+1 has no real roots (we are familiar with its graph, which is entirely above the xx-axis), it is irreducible, so now our denominator is fully factored.

x3+3(x2−1)2(x2+1)=x3+3(x−1)2(x+1)2(x2+1)=Ax−1+B(x−1)2+Cx+1+D(x+1)2+Ex+Fx2+1\begin{align*} \frac{x^3+3}{(x^2-1)^2(x^2+1)} &=\frac{x^3+3}{(x-1)^2(x+1)^2(x^2+1)}\\ &=\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{Ex+F}{x^2+1} \end{align*}

Notice (x−1)(x-1) and (x+1)(x+1) are (repeated) linear factors, while (x2+1)(x^2+1) is an irreducible quadratic factor. This accounts for the difference in the numerators of their corresponding terms.

Q3Stage 1Past exam · 2016Q4

Find the coefficient of 1x−1\displaystyle \frac{1}{x-1} in the partial fraction decomposition of 3x3−2x2+11x2(x−1)(x2+3)\displaystyle\frac{3x^3-2x^2+11}{x^2(x-1)(x^2+3)}.

Hint

Review Example 1.10.1 in the CLP-2 text. Is the “Algebraic Method” or the “Sneaky Method” going to be easier?

Answer

33

Full solution

The partial fraction decomposition has the form

3x3−2x2+11x2(x−1)(x2+3)=Ax−1+various terms\begin{align*} \frac{3x^3-2x^2+11}{x^2(x-1)(x^2+3)} = \frac A{x-1} + \text{various terms} \end{align*}

When we multiply through by the original denominator, this becomes

3x3−2x2+11=x2(x2+3)A+(x−1)(other terms).\begin{align*} {3x^3-2x^2+11} = {x^2(x^2+3)}A + (x-1)(\text{other terms}). \end{align*}

Evaluating both sides at x=1x=1 yields 3⋅13−2⋅12+11=12(12+3)A+03\cdot1^3-2\cdot1^2+11=1^2(1^2+3)A+0, or A=3A=3.

Q4Stage 1

Re-write the following rational functions as the sum of a polynomial and a rational function whose numerator has a strictly smaller degree than its denominator. (Remember our method of partial fraction decomposition of a rational function only works when the degree of the numerator is strictly smaller than the degree of the denominator.)

  1. x3+2x+2x2+1\dfrac{x^3+2x+2}{x^2+1}

  2. 15x4+6x3+34x2+4x+205x2+2x+8\dfrac{15x^4+6x^3+34x^2+4x+20}{5x^2+2x+8}

  3. 2x5+9x3+12x2+10x+302x2+5\dfrac{2x^5+9x^3+12x^2+10x+30}{2x^2+5}

Hint

For each part, use long division as in Example 1.10.4 of the CLP-2 text.

Answer

(a) x3+2x+2x2+1=x+x+2x2+1\displaystyle\frac{x^3+2x+2}{x^2+1} = x+\frac{x+2}{x^2+1}

(b) 15x4+6x3+34x2+4x+205x2+2x+8=3x2+2+45x2+2x+8\displaystyle\dfrac{15x^4+6x^3+34x^2+4x+20}{5x^2+2x+8} = 3x^2+2+\frac{4}{5x^2+2x+8}

(c) 2x5+9x3+12x2+10x+302x2+5=x3+2x+6\displaystyle\dfrac{2x^5+9x^3+12x^2+10x+30}{2x^2+5}=x^3+2x+6

Full solution

  1. We start by dividing. The leading term of the numerator is xx times the leading term of the denominator. The remainder is x+2x+2.

    Figure from prob_s1.10, line 242

    Figure from prob_s1.10, line 242

    That is, x3+2x+2=x(x2+1)+(x+2)x^3+2x+2 = x(x^2+1)+(x+2). So,

    x3+2x+2x2+1=x+x+2x2+1\frac{x^3+2x+2}{x^2+1} = x+\frac{x+2}{x^2+1}

  2. We start by dividing. The leading term of the numerator is 3x23x^2 times the leading term of the denominator.

    Figure from prob_s1.10, line 242

    Figure from prob_s1.10, line 242

    Then 5x25x^2 goes into 10x210x^2 twice, so

    Figure from prob_s1.10, line 242

    Figure from prob_s1.10, line 242

    Our remainder is 4. That is,

    15x4+6x3+34x2+4x+205x2+2x+8=3x2+2+45x2+2x+8.\dfrac{15x^4+6x^3+34x^2+4x+20}{5x^2+2x+8} = 3x^2+2+\frac{4}{5x^2+2x+8}.

  3. We start by dividing. The leading term of the numerator is x3x^3 times the leading term of the denominator.

    Figure from prob_s1.10, line 242

    Figure from prob_s1.10, line 242

    Then 2x2(2x)2x^2(2x) gives us 4x34x^3.

    Figure from prob_s1.10, line 242

    Figure from prob_s1.10, line 242

    Finally, 2x22x^2 goes into 12x212x^2 six times.

    Figure from prob_s1.10, line 242

    Figure from prob_s1.10, line 242

    Since there is no remainder,

    2x5+9x3+12x2+10x+302x2+5=x3+2x+6\dfrac{2x^5+9x^3+12x^2+10x+30}{2x^2+5}=x^3+2x+6

    Remark: if we wanted to be pedantic about the question statement, we could write our final answer as x3+2x+6+0xx^3+2x+6+\frac{0}{x}, so that we are indeed adding a polynomial to a rational function whose numerator has degree strictly smaller than its denominator.

Q5Stage 1

Factor the following polynomials into linear and irreducible factors.

  1. 5x3−3x2−10x+65x^3-3x^2-10x+6

  2. x4−3x2−5x^4-3x^2-5

  3. x4−4x3−10x2−11x−6x^4-4x^3-10x^2-11x-6

  4. 2x4+12x3−x2−52x+152x^4+12x^3-x^2-52x+15

Hint

(a) Look for a pattern you can exploit to factor out a linear term.
(b) If you set y=x2y=x^2, this is quadratic. Remember (x2−a)=(x+a)(x−a)(x^2-a)= (x+\sqrt{a})(x-\sqrt{a}) as long as aa is positive.
(c),(d) Look for integer roots, then use long division.

Answer

(a) 5x3−3x2−10x+6=(x+2)(x−2)(5x−3)5x^3-3x^2-10x+6=(x+\sqrt{2})(x-\sqrt{2})(5x-3)

(b) x4−3x2−5=(x+3+292)(x−3+292)(x2+29−32)x^4-3x^2-5=\displaystyle\left(x+\sqrt{\frac{3+\sqrt{29}}{2}}\right)\left(x-\sqrt{\frac{3+\sqrt{29}}{2}}\right)\left(x^2+\frac{\sqrt{29}-3}{2}\right)

(c) x4−4x3−10x2−11x−6=(x+1)(x−6)(x2+x+1)x^4-4x^3-10x^2-11x-6 = (x+1)(x-6)(x^2+x+1)
(d) 2x4+12x3−x2−52x+15=(x+3)(x+5)(x−(1+22))(x−(1−22))2x^4+12x^3-x^2-52x+15= (x+3)(x+5) \left(x-\Big(1+\frac{\sqrt2}{2}\Big)\right) \left(x-\Big(1-\frac{\sqrt2}{2}\Big)\right)

Full solution

  1. The polynomial 5x3−3x2−10x+65x^3-3x^2-10x+6 has a repeated pattern: the ratio of the first two coefficients is the same as the ratio of the last two coefficients. We can use this to factor.

    5x3−3x2−10x+6=x2(5x−3)−2(5x−3)=(x2−2)(5x−3)=(x+2)(x−2)(5x−3)\begin{align*} 5x^3-3x^2-10x+6&=x^2(5x-3)-2(5x-3) = (x^2-2)(5x-3)\\ &=(x+\sqrt{2})(x-\sqrt{2})(5x-3) \end{align*}

  2. The polynomial x4−3x2−5x^4-3x^2-5 has only even powers of xx, so we can (temporarily) replace them with x2=yx^2=y to turn our quartic polynomial into a quadratic.

    x4−3x2−5=y2−3y−5\begin{align*}x^4-3x^2-5&=y^2-3y-5\end{align*}

    There's no obvious factoring here, but we can find its roots, if any, using the quadratic equation.

    y=3±32−4(1)(−5)2=3±292So,y2−3y−5=(y−3+292)(y−3−292)Therefore,x4−3x2−5=(x2−3+292)(x2−3−292)\begin{align*}y&=\frac{3\pm\sqrt{3^2-4(1)(-5)}}{2}\\ &=\frac{3\pm\sqrt{29}}{2}\\ \text{So,}\qquad y^2-3y-5&=\left(y-\frac{3+\sqrt{29}}{2}\right)\left(y-\frac{3-\sqrt{29}}{2}\right)\\ \text{Therefore,}\qquad x^4-3x^2-5&=\left(x^2-\frac{3+\sqrt{29}}{2}\right)\left(x^2-\frac{3-\sqrt{29}}{2}\right)\end{align*}

    We'd like to use the difference of two squares to factor these quadratic expressions. For this to work, the constants must be positive (so their square roots are real). Since 29>3\sqrt{29}>3, only the first quadratic is factorable. The other is irreducible–it's always positive, so it had no roots.

    x4−3x2−5=(x+3+292)(x−3+292)(x2+29−32)\begin{align*}x^4-3x^2-5&=\left(x+\sqrt{\frac{3+\sqrt{29}}{2}}\right)\left(x-\sqrt{\frac{3+\sqrt{29}}{2}}\right)\left(x^2+\frac{\sqrt{29}-3}{2}\right)\end{align*}

  3. Without seeing any obvious patterns, we start hunting for roots. Since we have all integer coefficients, if there are any integer roots, they will divide our constant term, −6-6. So, our candidates for roots are ±1, ±2, ±3,\pm1,\,\pm2,\,\pm3, and ±6\pm6. To save time, we don't need to know exactly the value of our polynomial at these points: only whether or not it is 0. Write f(x)=x4−4x3−10x2−11x−6f(x) = x^4 - 4x^3 - 10x^2 - 11x - 6.

    f(−1)=0f(−2)≠0f(−3)≠0f(−6)≠0f(1)≠0f(2)≠0f(3)≠0f(6)=0\begin{align*} \color{red}f(-1)&\color{red}=0& f(-2)&\neq 0& f(-3)&\neq 0& f(-6)&\neq 0\\ f(1)&\neq 0& f(2)&\neq 0& f(3)&\neq 0& \color{red}f(6)&\color{red}= 0 \end{align*}

    Since x=−1x=-1 and x=6x=6 are roots of our polynomial, it has factors (x+1)(x+1) and (x−6)(x-6). Note (x+1)(x−6)=x2−5x−6(x+1)(x-6) = x^2-5x-6. We use long division to figure out what else is lurking in our polynomial.

    Figure from prob_s1.10, line 306

    Figure from prob_s1.10, line 306

    So, x4−4x3−10x2−11x−6=(x+1)(x−6)(x2+x+1)x^4-4x^3-10x^2-11x-6 = (x+1)(x-6)(x^2+x+1).

    We should check whether x2+x+1x^2+x+1 is reducible or not. If we try to find its roots with the quadratic equation, we get −1±−32\dfrac{-1\pm\sqrt{-3}}{2}, which are not real numbers. So, we're at the end of our factoring.

  4. Without seeing any obvious patterns, we start hunting for roots. Since we have all integer coefficients, if there are any integer roots, they will divide our constant term, −15-15. So, our candidates for roots are ±1, ±3, ±5,\pm1,\,\pm3,\,\pm5, and ±15\pm15. Write f(x)=2x4+12x3−x2−52x+15f(x) = 2x^4 + 12x^3 - x^2 - 52x + 15.

    f(−1)≠0f(−3)=0f(−5)=0f(−5)≠0f(1)≠0f(3)≠0f(5)≠0f(15)≠0\begin{align*} f(-1)&\neq 0& \color{red}f(-3)&\color{red}=0& \color{red}f(-5)&\color{red}= 0& f(-5)&\neq 0\\ f(1)&\neq 0& f(3)&\neq 0& f(5)&\neq 0& f(15)&\neq 0& \end{align*}

    Since x=−3x=-3 and x=−5x=-5 are roots of our polynomial, it has factors (x+3)(x+3) and (x+5)(x+5). Note (x+3)(x+5)=x2+8x+15(x+3)(x+5) = x^2+8x+15. We use long division to move forward.

    Figure from prob_s1.10, line 306

    Figure from prob_s1.10, line 306

    So, 2x4+12x3−x2−52x+15=(x+3)(x+5)(2x2−4x+1)2x^4+12x^3-x^2-52x+15= (x+3)(x+5)(2x^2-4x+1).

    We should check whether 2x2−4x+12x^2-4x+1 is reducible or not. There's not an obvious way to factor it, but we can use the quadratic equation. This gives us roots 4±16−84=1±22\dfrac{4\pm\sqrt{16-8}}{4}=1\pm\frac{\sqrt2}{2}. So, we have two more linear factors.

    Specifically: 2x4+12x3−x2−52x+15=(x+3)(x+5)(x−(1+22))(x−(1−22))2x^4+12x^3-x^2-52x+15= (x+3)(x+5)\left(x-\Big(1+\frac{\sqrt2}{2}\Big)\right) \left(x-\Big(1-\frac{\sqrt2}{2}\Big)\right).

Q6Stage 1

Here is a fact:

Suppose we have a rational function with a repeated linear factor (ax+b)n(ax+b)^n in the denominator, and the degree of the numerator is strictly less than the degree of the denominator. In the partial fraction decomposition, we can replace the terms

A1ax+b+A2(ax+b)2+A3(ax+b)3+⋯+An(ax+b)n\begin{equation}\frac{A_1}{ax+b} + \frac{A_2}{(ax+b)^2}+\frac{A_3}{(ax+b)^3}+\cdots+ \frac{A_n}{(ax+b)^n}\tag{1}\end{equation}

with the single term

B1+B2x+B3x2+⋯+Bnxn−1(ax+b)n\begin{equation}\frac{B_1+B_2x+B_3x^2+\cdots + B_{n}x^{n-1}}{(ax+b)^n}\tag{2}\end{equation}

and still be guaranteed to find a solution.

Why do we use the sum in (1), rather than the single term in (2), in partial fraction decomposition?

Hint

Why do we do partial fraction decomposition at all?

Answer

The goal of partial fraction decomposition is to write our integrand in a form that is easy to integrate. The antiderivative of (1) can be easily determined with the substitution u=(ax+b)u=(ax+b). It's less clear how to find the antiderivative of (2).

Full solution

The goal of partial fraction decomposition is to write our integrand in a form that is easy to integrate. The antiderivative of (1) can be easily determined with the substitution u=(ax+b)u=(ax+b). It's less clear how to find the antiderivative of (2).

2Stage 2Procedural

Practising the skill itself, until applying it is automatic.

Q7Stage 2Past exam · 2013A

Evaluate ∫12dxx+x2\displaystyle\int_1^2 \frac{\dee{x}}{x+x^2}.

Hint

What is the title of this section?

Answer

log⁡43\displaystyle\log\frac{4}{3}

Full solution

The integrand is a rational function, so it's a candidate for partial fraction. We quickly rule out any obvious substitution or integration by parts, so we go ahead with the decomposition.

We start by expressing the integrand, i.e. the fraction 1x+x2=1x(1+x)\frac{1}{x+x^2}=\frac{1}{x(1+x)}, as a linear combination of the simpler fractions 1x\frac{1}{x} and 1x+1\frac{1}{x+1} (which we already know how to integrate). We will have

1x+x2=1x(1+x)=ax+bx+1=a(x+1)+bxx(1+x)\begin{align*} \frac{1}{x+x^2}=\frac{1}{x(1+x)} = \frac{a}{x} + \frac{b}{x+1} =\frac{a(x+1) +bx}{x(1+x)} \end{align*}

The fraction on the left hand side is the same as the fraction on the right hand side if and only if the numerator on the left hand side, which is 1=0x+11 = 0x + 1, is equal to the numerator on the right hand side, which is a(x+1)+bx=(a+b)x+aa(x+1)+bx = (a+b)x +a. This in turn is the case if and only of a=1a=1 (i.e. the constant terms are the same in the two numerators) and a+b=0a+b=0 (i.e. the coefficients of xx are the same in the two numerators). So a=1a=1 and b=−1b=-1. Now we can easily evaluate the integral

∫12dxx+x2=∫12dxx(x+1)=∫12(1x−1x+1) dx=[log⁡x−log⁡(x+1)]12=log⁡2−log⁡32=log⁡43\begin{align*} \int_1^2 \frac{\dee{x}}{x+x^2} &=\int_1^2 \frac{\dee{x}}{x(x+1)} =\int_1^2 \Big(\frac{1}{x}-\frac{1}{x+1}\Big)\,\dee{x} =\Big[\log x-\log(x+1)\Big]_1^2\\ &=\log 2-\log\frac{3}{2} =\log\frac{4}{3} \end{align*}
Q8Stage 2Past exam · 2015A

Calculate ∫1x4+x2 dx\displaystyle \int \frac{1}{x^4+x^2}\,\dee{x}.

Hint

You can save yourself some work in developing your partial fraction decomposition by renaming x2x^2 to yy and comparing the result with Question 7.

Answer

−1x−arctan⁡x+C-\dfrac{1}{x}-\arctan x+C

Full solution

We'll first do a partial fraction decomposition. The sneaky way is to temporarily rename x2x^2 to yy. Then x4+x2=y2+yx^4+x^2=y^2+y and

1x4+x2=1y(y+1)=1y−1y+1\begin{equation*} \frac{1}{x^4+x^2} = \frac{1}{y(y+1)} =\frac{1}{y}-\frac{1}{y+1} \end{equation*}

as we found in Question 7. Now we restore yy to x2x^2.

∫1x4+x2 dx=∫(1x2−1x2+1) dx=−1x−arctan⁡x+C\begin{align*} \int \frac1{x^4+x^2}\,\dee{x} =\int \Big(\frac{1}{x^2}-\frac{1}{x^2+1}\Big)\,\dee{x} =-\frac{1}{x}-\arctan x+C \end{align*}
Q9Stage 2Past exam · 2016A

Calculate ∫12x+4(x−3)(x2+1) dx\displaystyle \int \frac{12x+4}{(x-3)(x^2+1)}\,\dee{x}.

Hint

Review Steps 3 (particularly the “Sneaky Method”) and 4 of Example 1.10.3 in the CLP-2 text.

Answer

4log⁡∣x−3∣−2log⁡(x2+1)+C4 \log |x-3| - 2 \log (x^2 + 1) + C

Full solution

The integrand is of the form N(x)/D(x)N(x)/D(x) with D(x)D(x) already factored and N(x)N(x) of lower degree. We immediately look for a partial fraction decomposition:

12x+4(x−3)(x2+1)=Ax−3+Bx+Cx2+1.\begin{equation*} \frac{12x+4}{(x-3)(x^2+1)} = \frac{A}{x-3} + \frac{Bx+C}{x^2+1}. \end{equation*}

Multiplying through by the denominator yields

12x+4=A(x2+1)+(Bx+C)(x−3)\begin{align} 12x+4 &= A(x^2+1) + (Bx+C)(x-3) \tag{$*$} \end{align}

Setting x=3x=3 we find:

36+4=A(9+1)+0  ⟹  40=10A  ⟹  A=4\begin{align*} 36 +4 = A(9 + 1) + 0 \implies 40 =10A \implies \color{red}A = 4 \end{align*}

Substituting A=4A=4 in (∗)(*) gives

12x+4=4(x2+1)+(Bx+C)(x−3)  ⟹  −4x2+12x=(x−3)(Bx+C)  ⟹  (−4x)(x−3)=(Bx+C)(x−3)  ⟹  B=−4, C=0\begin{align*} 12x+4 = 4(x^2+1) + (Bx+C)(x-3) &\implies -4x^2+12 x =(x-3)(Bx + C) \\ &\implies (-4x)(x-3)=(Bx + C)(x-3) \\ & \implies \color{red}B=-4,~C=0 \end{align*}

So we have found that A=4A=4, B=−4B=-4, and C=0C=0. Therefore

∫12x+4(x−3)(x2+1) dx=∫(4x−3−4xx2+1) dx=4log⁡∣x−3∣−2log⁡(x2+1)+C\begin{align*} \int \frac{12x+4}{(x-3)(x^2+1)} \,\dee{x} &= \int \bigg( \frac{4}{x-3} - \frac{4x}{x^2+1} \bigg) \,\dee{x} \\ &= 4 \log |x-3| - 2 \log (x^2 + 1) + C \end{align*}

The second integral was found just by guessing an antiderivative. Alternatively, one could use the substitution u=x2+1u=x^2+1, du=2x dx\dee{u}=2x\,\dee{x}.

Q10Stage 2Past exam · 2016Q4

Evaluate the following indefinite integral using partial fraction:

F(x)=∫3x2−4(x−2)(x2+4) dx.\begin{align*} F(x) = \int \frac{3x^2 -4}{(x-2)(x^2+4)}\,\dee{x} . \end{align*}
Hint

Review Steps 3 (particularly the “Sneaky Method”) and 4 of Example 1.10.3 in the CLP-2 text. Remember ddx{arctan⁡x}=11+x2\diff{}{x}\{\arctan x\} = \frac{1}{1+x^2}.

Answer

F(x)=log⁡∣x−2∣+log⁡(x2+4)+2arctan⁡(x/2)+DF(x) = \log |x-2| + \log (x^2+4) + 2\arctan (x/2) + D

Full solution

The integrand is of the form N(x)/D(x)N(x)/D(x) with D(x)D(x) already factored and N(x)N(x) of lower degree. With no obvious substitution available, we look for a partial fraction decomposition.

3x2−4(x−2)(x2+4)=Ax−2+Bx+Cx2+4\begin{align*} \frac{3x^2 -4}{(x-2)(x^2+4)} = \frac{A}{x-2} + \frac{Bx + C}{x^2+4} \end{align*}

Multiplying through by the denominator gives

3x2−4=A(x2+4)+(Bx+C)(x−2)\begin{align} 3x^2 -4 = A(x^2+4) + (Bx + C)(x-2) \tag{$*$} \end{align}

Setting x=2x=2 we find:

12−4=A(4+4)+0  ⟹  8=8A  ⟹  A=1\begin{align*} 12 -4 = A(4 + 4) + 0 \implies 8 =8A \implies \color{red}A = 1 \end{align*}

Substituting A=1A=1 in (∗)(*) gives

3x2−4=(x2+4)+(x−2)(Bx+C)  ⟹  2x2−8=(x−2)(Bx+C)  ⟹  (x−2)(2x+4)=(x−2)(Bx+C)  ⟹  B=2, C=4\begin{align*} 3x^2 -4 = (x^2+4) + (x-2)(Bx + C) &\implies 2x^2-8 =(x-2)(Bx + C) \\ &\implies (x-2)(2x+4)=(x-2)(Bx + C) \\ & \implies \color{red} B=2,~C=4 \end{align*}

Thus, we have:

3x2−4(x−2)(x2+4)=1x−2+2x+4x2+4=1x−2+2xx2+4+4x2+4\begin{align*} \frac{3x^2 -4}{(x-2)(x^2+4)} = \frac{1}{x-2} + \frac{2x + 4}{x^2+4} = \frac{1}{x-2} + \frac{2x }{x^2+4} + \frac{4}{x^2+4 } \end{align*}

The first two of these are directly integrable:

F(x)=log⁡∣x−2∣+log⁡∣x2+4∣+∫4x2+4 dx\begin{align*} F(x) = \log |x-2| + \log |x^2+4| + \int \frac{4}{x^2+4}\,\dee{x} \end{align*}

(The second integral was found just by guessing an antiderivative. Alternatively, one could use the substitution u=x2+4u=x^2+4, du=2x dx\dee{u}=2x\,\dee{x}.) For the final integral, we substitute: x=2yx = 2y, dx=2dy\dee{x} = 2\dee{y}, and see that:

∫4x2+4 dx=2∫1y2+1 dy=2arctan⁡y+D=2arctan⁡(x/2)+D\begin{align*} \int \frac{4}{x^2+4}\,\dee{x} = 2 \int \frac{1}{y^2 + 1}\,\dee{y} = 2 \arctan y + D = 2\arctan (x/2) + D \end{align*}

for any constant DD. All together we have:

F(x)=log⁡∣x−2∣+log⁡(x2+4)+2arctan⁡(x/2)+D\begin{align*} F(x) = \log |x-2| + \log (x^2+4) + 2\arctan (x/2) + D \end{align*}
Q11Stage 2Past exam · M105 2014A

Evaluate ∫x−13x2−x−6dx\displaystyle\int \frac{x-13}{x^2-x-6}\dee{x}.

Hint

Fill in the blank: the integrand is a         ‾\underbar{\ \ \ \ \ \ \ \ } function.

Answer

−2log⁡∣x−3∣+3log⁡∣x+2∣+C-2\log|x-3|+3\log|x+2|+C

Full solution

This integrand is a rational function, with no obvious substitution. This sure looks like a partial fraction problem. Let's go through our protocol.

  • The degree of the numerator x−13x-13 is one, which is strictly smaller than the dergee of the denominator x2−x−6x^2-x-6, which is two. So we don't need long division to pull out a polynomial.

  • Next we factor the denominator.

    x2−x−6=(x−3)(x+2)\begin{align*} x^2-x-6 = (x-3)(x+2) \end{align*}

  • Next we find the partial fraction decomposition of the integrand. It is of the form

    x−13(x−3)(x+2)=Ax−3+Bx+2\begin{align*} \frac{x-13}{(x-3)(x+2)} =\frac{A}{x-3} + \frac{B}{x+2} \end{align*}

    To find AA and BB, using the sneaky method, we cross multiply by the denominator.

    x−13=A(x+2)+B(x−3)\begin{align*} x-13 = A(x+2) + B(x-3) \end{align*}

    Now we can find AA by evaluating at x=3x=3

    3−13=A(3+2)+B(3−3)  ⟹  A=−2\begin{align*} 3-13 = A(3+2) + B(3-3) \implies \color{red}A=-2 \end{align*}

    and find BB by evaluating at x=−2x=-2.

    −2−13=A(−2+2)+B(−2−3)  ⟹  B=3\begin{align*} -2-13 = A(-2+2) + B(-2-3) \implies \color{red}B=3 \end{align*}

    (Hmmm. AA and BB are nice round numbers. Sure looks like a rigged exam or homework question.) Our partial fraction decomposition is

    x−13(x−3)(x+2)=−2x−3+3x+2\begin{align*} \frac{x-13}{(x-3)(x+2)} =\frac{-2}{x-3} + \frac{3}{x+2} \end{align*}

    As a check, we recombine the right hand side and make sure that it matches the left hand side.

    −2x−3+3x+2=−2(x+2)+3(x−3)(x−3)(x+2)=x−13(x−3)(x+2)\begin{align*} \frac{-2}{x-3} + \frac{3}{x+2} =\frac{-2(x+2)+3(x-3)}{(x-3)(x+2)} =\frac{x-13}{(x-3)(x+2)} \end{align*}

  • Finally, we evaluate the integral.

    ∫x−13x2−x−6dx=∫(−2x−3+3x+2)dx=−2log⁡∣x−3∣+3log⁡∣x+2∣+C\begin{align*} \int \frac{x-13}{x^2-x-6}\dee{x} =\int\bigg(\frac{-2}{x-3} + \frac{3}{x+2}\bigg)\dee{x} =-2\log|x-3|+3\log|x+2|+C \end{align*}
Q12Stage 2Past exam · 2014D

Evaluate ∫5x+1x2+5x+6dx\displaystyle\int \frac{5x+1}{x^2+5x+6}\dee{x}.

Hint

The integrand is yet another         ‾\underbar{\ \ \ \ \ \ \ \ } function.

Answer

−9log⁡∣x+2∣+14log⁡∣x+3∣+C-9\log|x+2|+14\log|x+3| +C

Full solution

Again, this sure looks like a partial fraction problem. So let's go through our protocol.

  • The degree of the numerator 5x+15x+1 is one, which is strictly smaller than the dergee of the denominator x2+5x+6x^2+5x+6, which is two. So we do not long divide to pull out a polynomial.

  • Next we factor the denominator.

    x2+5x+6=(x+2)(x+3)\begin{align*} x^2+5x+6 = (x+2)(x+3) \end{align*}

  • Next we find the partial fraction decomposition of the integrand. It is of the form

    5x+1(x+2)(x+3)=Ax+2+Bx+3\begin{align*} \frac{5x+1}{(x+2)(x+3)} =\frac{A}{x+2} + \frac{B}{x+3} \end{align*}

    To find AA and BB, using the sneaky method, we cross multiply by the denominator.

    5x+1=A(x+3)+B(x+2)\begin{align*} 5x+1= A(x+3) + B(x+2) \end{align*}

    Now we can find AA by evaluating at x=−2x=-2

    −10+1=A(−2+3)+B(−2+2)  ⟹  A=−9\begin{align*} -10+1 = A(-2+3) + B(-2+2) \implies\color{red} A=-9 \end{align*}

    and find BB by evaluating at x=−3x=-3.

    −15+1=A(−3+3)+B(−3+2)  ⟹  B=14\begin{align*} -15+1 = A(-3+3) + B(-3+2) \implies\color{red} B=14 \end{align*}

    So our partial fraction decomposition is

    5x+1(x+2)(x+3)=−9x+2+14x+3\begin{align*} \frac{5x+1}{(x+2)(x+3)} =\frac{-9}{x+2} + \frac{14}{x+3} \end{align*}

    As a check, we recombine the right hand side and make sure that it matches the left hand side

    −9x+2+14x+3=−9(x+3)+14(x+2)(x+2)(x+3)=5x+1(x+2)(x+3)\begin{align*} \frac{-9}{x+2} + \frac{14}{x+3} =\frac{-9(x+3)+14(x+2)}{(x+2)(x+3)} =\frac{5x+1}{(x+2)(x+3)} \end{align*}

  • Finally, we evaluate the integral

    ∫5x+1x2+5x+6dx=∫(−9x+2+14x+3)dx=−9log⁡∣x+2∣+14log⁡∣x+3∣+C\begin{align*} \int \frac{5x+1}{x^2+5x+6}\dee{x} =\int\bigg(\frac{-9}{x+2} + \frac{14}{x+3}\bigg)\dee{x} =-9\log|x+2|+14\log|x+3|+C \end{align*}
Q13Stage 2

Evaluate ∫5x2−3x−1x2−1 dx\displaystyle\int \frac{5x^2-3x-1}{x^2-1}~\dee{x}.

Hint

Since the degree of the numerator is the same as the degree of the denominator, we can't do our partial fraction decomposition before we simplify the integrand.

Answer

5x+12log⁡∣x−1∣−72log⁡∣x+1∣+C\displaystyle5x+\frac{1}{2}\log|x-1| - \frac{7}{2}\log|x+1|+C

Full solution

We have a rational function with no obvious substitution, so let's use partial fraction decomposition.

  • Since the degree of the numerator is the same as the degree of the denominator, we need to pull out a polynomial.

    Figure from prob_s1.10, line 756

    Figure from prob_s1.10, line 756

    That is,

    ∫5x2−3x−1x2−1dx=∫(5+−3x+4x2−1)dx=5x+∫−3x+4x2−1dx\int \frac{5x^2-3x-1}{x^2-1}\dee{x} = \int\left(5+ \frac{-3x+4}{x^2-1}\right)\dee{x} =5x+ \int \frac{-3x+4}{x^2-1}\dee{x}

  • Again, there's no obvious substitution for the new integrand, so we want to use partial fraction. The denominator factors as (x−1)(x+1)(x-1)(x+1), so our decomposition has this form:

    −3x+4x2−1=−3x+4(x−1)(x+1)=Ax−1+Bx+1=(A+B)x+(A−B)(x−1)(x+1)\begin{align*} \frac{-3x+4}{x^2-1}&=\frac{-3x+4}{(x-1)(x+1)} = \frac{A}{x-1}+\frac{B}{x+1} = \frac{(A+B)x+(A-B)}{(x-1)(x+1)} \end{align*}

    So, (1) A+B=−3A+B=-3 and (2) A−B=4A-B=4.

  • We solve (2) for AA in terms of BB, namely A=4+BA=4+B. Plugging this into (1), we see (4+B)+B=−3(4+B)+B=-3. So, B=−72B=-\frac{7}{2}, and A=12A=\frac{1}{2}.

  • Now we can write our integral in a friendlier form and evaluate.

    ∫5x2−3x−1x2−1dx==5x+∫−3x+4x2−1dx=5x+∫1/2x−1−7/2x+1 dx=5x+12log⁡∣x−1∣−72log⁡∣x+1∣+C\begin{align*} \int \frac{5x^2-3x-1}{x^2-1}\dee{x} =&=5x+ \int \frac{-3x+4}{x^2-1}\dee{x} = 5x+\int \frac{1/2}{x-1} - \frac{7/2}{x+1}~\dee{x}\\ &=5x+\frac{1}{2}\log|x-1| - \frac{7}{2}\log|x+1|+C \end{align*}
Q14Stage 2

Evaluate ∫4x4+14x2+24x4+x2 dx\displaystyle\int \frac{4x^4+14x^2+2}{4x^4+x^2}~\dee{x}.

Hint

The degree of the numerator is not smaller than the degree of the denominator.
Your final answer will have an arctangent in it.

Answer

x−2x+52arctan⁡(2x)+C\displaystyle x-\frac{2}{x}+\frac{5}{2}\arctan (2x) +C

Full solution

The integrand is a rational function with no obvious substitution, so we use partial fraction decomposition.

  • The degree of the numerator is the same as the degree of the denominator. Since it's not smaller, we need to re-write our integrand. We could do this using long division, but this case is simple enough to do more informally.

    4x4+14x2+24x4+x2=4x4+x2+13x2+24x4+x2=4x4+x24x4+x2+13x2+24x4+x2=1+13x2+24x4+x2\begin{align*} \frac{4x^4+14x^2+2}{4x^4+x^2}&=\frac{4x^4+x^2+13x^2+2}{4x^4+x^2}\\ &=\frac{4x^4+x^2}{4x^4+x^2}+\frac{13x^2+2}{4x^4+x^2} \\&=1+\frac{13x^2+2}{4x^4+x^2} \end{align*}

  • The denominator factors as x2(4x2+1)x^2(4x^2+1).

  • We want to find the partial fraction decomposition of the fractional part of our simplified integrand.

    13x2+24x4+x2=13x2+2x2(4x2+1)=Ax+Bx2+Cx+D4x2+1\begin{align*}\frac{13x^2+2}{4x^4+x^2}&= \frac{13x^2+2}{x^2(4x^2+1)} = \frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{4x^2+1}\end{align*}

    Multiply through by the original denominator.

    13x2+2=Ax(4x2+1)+B(4x2+1)+(Cx+D)x2\begin{align*}13x^2+2&=Ax(4x^2+1)+B(4x^2+1)+(Cx+D)x^2 \tag{1}\end{align*}

    Setting x=0x=0 gives us:

    2=B\begin{align*}\textcolor{red}{2}&\color{red}=B\end{align*}

    We use B=2B=2 to simplify Equation (1).

    13x2+2=Ax(4x2+1)+2(4x2+1)+(Cx+D)x25x2=Ax(4x2+1)+(Cx+D)x25x=A(4x2+1)+(Cx+D)x\begin{align*}13x^2+2&=Ax(4x^2+1)+\textcolor{red}{2}(4x^2+1)+(Cx+D)x^2\\ 5x^2&=Ax(4x^2+1)+(Cx+D)x^2\\ 5x&=A(4x^2+1)+(Cx+D)x\tag{2}\end{align*}

    Again, let x=0x=0.

    0=A\begin{align*}\color{red}{0}&\color{red}=A\end{align*}

    Using A=0A=0, simplify Equation (2).

    5x=(Cx+D)x5=Cx+DC=0,D=5\begin{align*}5x&=(Cx+D)x\\ 5&=Cx+D\\ \color{red}C& \textcolor{red}{=0},\qquad\color{red}{D=5}\end{align*}

  • Now we can write our integral in pieces.

    ∫4x4+14x2+24x4+x2 dx=∫(1+13x2+24x4+x2) dx=∫(1+2x2+54x2+1) dx=x−2x+∫5(2x)2+1 dx\begin{align*}\int \frac{4x^4+14x^2+2}{4x^4+x^2}~\dee{x}&= \int \left(1+\frac{13x^2+2}{4x^4+x^2}\right)~\dee{x}\\ &=\int\left(1+ \frac{\textcolor{red}2}{x^2}+\frac{\textcolor{red}5}{4x^2+1}\right)~\dee{x} \\&=x -\frac{2}{x} + \int \frac{5}{(2x)^2+1} ~\dee{x}\end{align*}

    Substitute u=2xu=2x, du=2dx\dee{u}=2\dee{x}.

    =x−2x+∫5/2u2+1 du=x−2x+52arctan⁡u+C=x−2x+52arctan⁡(2x)+C\begin{align*}& =x -\frac{2}{x} + \int \frac{5/2}{u^2+1}~\dee{u} \\ &=x-\frac{2}{x}+\frac{5}{2}\arctan u +C\\ &=x-\frac{2}{x}+\frac{5}{2}\arctan (2x) +C\end{align*}
Q15Stage 2

Evaluate ∫x2+2x−1x4−2x3+x2 dx\displaystyle\int \frac{x^2+2x-1}{x^4-2x^3+x^2}~\dee{x}.

Hint

In the partial fraction decomposition, several constants turn out to be 0.

Answer

1x−2x−1+C\displaystyle\frac{1}{x}-\frac{2}{x-1}+C

Full solution

The integrand is a rational function with no obvious substitution, so we'll use a partial fraction decomposition.

  • Since the numerator has strictly smaller degree than the denominator, we don't need to start off with a long division.

  • We do, however, need to factor the denominator. We can immediately pull out x2x^2; the remaining part is x2−2x+1=(x−1)2x^2-2x+1 = (x-1)^2.

  • Now we can perform our partial fraction decomposition.

    x2+2x−1x4−2x3+x2=x2+2x−1x2(x−1)2=Ax+Bx2+Cx−1+D(x−1)2\begin{align*}\frac{x^2+2x-1}{x^4-2x^3+x^2}&= \frac{x^2+2x-1}{x^2(x-1)^2} = \frac{A}{x}+\frac{B}{x^2}+\frac{C}{x-1}+\frac{D}{(x-1)^2}\end{align*}

    Multiply both sides by the original denominator.

    x2+2x−1=Ax(x−1)2+B(x−1)2+Cx2(x−1)+Dx2\begin{align*}x^2+2x-1&=Ax(x-1)^2+B(x-1)^2+Cx^2(x-1)+Dx^2\tag{1}\end{align*}

    To be sneaky, we set x=0x=0, and find:

    −1=B\begin{align*}\color{red}-1&\color{red}=B\end{align*}

    We also set x=1x=1, and find:

    2=D\begin{align*}\color{red}2&\color{red}=D\end{align*}

    We use BB and DD to simplify Equation (1).

    x2+2x−1=Ax(x−1)2−1(x−1)2+Cx2(x−1)+2x20=Ax(x−1)2+Cx2(x−1)=x(x−1)[(A+C)x−A]So,0=(A+C)x−A\begin{align*}x^2+2x-1&=Ax(x-1)^2\textcolor{red}{-1}(x-1)^2+Cx^2(x-1)+\textcolor{red}{2}x^2\\ 0&=Ax(x-1)^2+Cx^2(x-1)\\ &=x(x-1)[(A+C)x-A]\\ \text{So,}\qquad 0&=(A+C)x-A\end{align*}

    That is, A=C=0A=C=0.

  • Now we can evaluate our integral.

    ∫x2+2x−1x4−2x3+x2 dx=∫(−1x2+2(x−1)2) dx=1x−2x−1+C\begin{align*} \int \frac{x^2+2x-1}{x^4-2x^3+x^2}~\dee{x}&=\int \left(\frac{-1}{x^2}+\frac{2}{(x-1)^2}\right)~\dee{x}\\ &=\frac{1}{x}-\frac{2}{x-1}+C \end{align*}
Q16Stage 2

Evaluate ∫3x2−4x−102x3−x2−8x+4 dx\displaystyle\int \frac{ 3x^2-4x-10}{2x^3-x^2-8x+4}~\dee{x}.

Hint

Factor (2x−1)(2x-1) out of the denominator to get started. You don't need long division for this step.

Answer

−12log⁡∣x−2∣+12log⁡∣x+2∣+32log⁡∣2x−1∣+C\displaystyle-\frac{1}{2}\log|x-2| + \frac{1}{2}\log|x+2| + \frac{3}{2}\log|2x-1|+C

Full solution

Our integrand is a rational function with no obvious substitution, so we'll use the method of partial fractions.

  • The degree of the numerator is less than the degree of the denominator.

  • We need to factor the denominator. The first two terms have the same ratio as the last two terms.

    2x3−x2−8x+4=x2(2x−1)−4(2x−1)=(x2−4)(2x−1)=(x−2)(x+2)(2x−1)\begin{align*} 2x^3-x^2-8x+4&=x^2(2x-1)-4(2x-1) \\ &= (x^2-4)(2x-1)\\ &=(x-2)(x+2)(2x-1) \end{align*}

  • Now we find our partial fraction decomposition.

    3x2−4x−102x3−x2−8x+4=3x2−4x−10(x−2)(x+2)(2x−1)=Ax−2+Bx+2+C2x−1\begin{align*}\frac{ 3x^2-4x-10}{2x^3-x^2-8x+4}&=\frac{ 3x^2-4x-10}{(x-2)(x+2)(2x-1)}= \frac{A}{x-2}+\frac{B}{x+2}+\frac{C}{2x-1}\end{align*}

    Multiply both sides by the original denominator.

    3x2−4x−10=A(x+2)(2x−1)+B(x−2)(2x−1)+C(x−2)(x+2)\begin{align*}3x^2-4x-10&=A(x+2)(2x-1)+B(x-2)(2x-1)+C(x-2)(x+2)\end{align*}

    Distinct linear factors is the best possible scenario for the sneaky method. First, let's set x=2x=2.

    3(4)−4(2)−10=A(4)(3)+B(0)+C(0)A=−12\begin{align*}3(4)-4(2)-10&=A(4)(3)+B(0)+C(0)\\ \color{red}A&\color{red}=-\frac{1}{2}\end{align*}

    Now, let x=−2x=-2.

    3(4)−4(−2)−10=A(0)+B(−4)(−5)+C(0)B=12\begin{align*}3(4)-4(-2)-10&=A(0)+B(-4)(-5)+C(0)\\ \color{red}B&\color{red}=\frac{1}{2}\end{align*}

    Finally, let x=12x=\frac{1}{2}.

    34−2−10=A(0)+B(0)+C(−32)(52)C=3\begin{align*}\frac{3}{4}-2-10&=A(0)+B(0)+C\left(-\frac{3}{2}\right)\left(\frac{5}{2}\right)\\ \color{red}C&\color{red}=3\end{align*}

  • Now we can evaluate our integral in its new form.

    ∫3x2−4x−102x3−x2−8x+4 dx=∫(−1/2x−2+1/2x+2+32x−1) dx=−12log⁡∣x−2∣+12log⁡∣x+2∣+32log⁡∣2x−1∣+C=12log⁡∣x+2x−2∣+32log⁡∣2x−1∣+C\begin{align*} \int \frac{ 3x^2-4x-10}{2x^3-x^2-8x+4}~\dee{x}&=\int \left(\frac{-1/2}{x-2}+\frac{1/2}{x+2}+\frac{3}{2x-1}\right) ~\dee{x}\\ &=-\frac{1}{2}\log|x-2| + \frac{1}{2}\log|x+2| + \frac{3}{2}\log|2x-1|+C \\&=\frac{1}{2}\log\left| \frac{x+2}{x-2} \right| + \frac{3}{2}\log|2x-1|+C \end{align*}
Q17Stage 2

Evaluate ∫0110x2+24x+82x3+11x2+6x+5 dx\displaystyle\int_0^1 \frac{10x^2+24x+8}{2x^3+11x^2+6x+5}~\dee{x}.

Hint

When it comes time to integrate, look for a convenient substitution.

Answer

log⁡(4⋅6353)\displaystyle\log \left(\frac{4\cdot 6^3}{5^3}\right)

Full solution

The integrand is a rational function with no obvious substitution, so we use the method of partial fractions.

  • The numerator has smaller degree than the denominator.

  • We need to factor the denominator. In the absence of any clues, we look for an integer root. The constant term is 5, so the possible integer roots are ±1\pm 1 and ±5\pm 5. Name f(x)=2x3+11x2+6x+5f(x) = 2x^3+11x^2+6x+5.

    f(−1)≠0f(-1)\neq0 f(−5)=0f(-5)=0 f(1)≠0f(1)\neq 0 f(5)≠0f(5)\neq 0  

    So, (x+5)(x+5) is a factor of the denominator.

  • We use long division to pull out the factor of (x+5)(x+5).

    Figure from prob_s1.10, line 943

    Figure from prob_s1.10, line 943

    That is, our denominator is (x+5)(2x2+x+1)(x+5)(2x^2+x+1).

  • The quadratic function 2x2+x+12x^2+x+1 is irreducible: we can see this by using the quadratic equation, and finding no real roots. So, we are ready to find our partial fraction decomposition.

    10x2+24x+82x3+11x2+6x+5=10x2+24x+8(x+5)(2x2+x+1)=Ax+5+Bx+C2x2+x+1\begin{align*}\frac{10x^2+24x+8}{2x^3+11x^2+6x+5}&=\frac{10x^2+24x+8}{(x+5)(2x^2+x+1)}=\frac{A}{x+5}+\frac{Bx+C}{2x^2+x+1}\end{align*}

    Multiply through by the original denominator.

    10x2+24x+8=A(2x2+x+1)+(Bx+C)(x+5)\begin{align*}10x^2+24x+8&=A(2x^2+x+1)+(Bx+C)(x+5)\tag{1}\end{align*}

    Set x=−5x=-5.

    10(25)−24(5)+8=A(2(25)−5+1)+(B(−5)+C)(0)A=3\begin{align*}10(25)-24(5)+8&=A(2(25)-5+1) + (B(-5)+C)(0)\\ \color{red}A&\color{red}=3\end{align*}

    Using our value of AA, we simplify Equation (1).

    10x2+24x+8=3(2x2+x+1)+(Bx+C)(x+5)4x2+21x+5=(Bx+C)(x+5)\begin{align*}10x^2+24x+8&=\textcolor{red}{3}(2x^2+x+1)+(Bx+C)(x+5) \\4x^2+21x+5&=(Bx+C)(x+5)\end{align*}

    We factor the left side. We know (x+5)(x+5) must be one of its factors.

    (4x+1)(x+5)=(Bx+C)(x+5)4x+1=Bx+C\begin{align*}(4x+1)(x+5)&=(Bx+C)(x+5)\\ 4x+1&=Bx+C\end{align*}

    So, B=4\textcolor{red}{B=4} and C=1\textcolor{red}{C=1}.

  • Now we can write our integral in smaller pieces.

    ∫0110x2+24x+82x3+11x2+6x+5 dx=∫01(3x+5+4x+12x2+x+1) dx\begin{align*}\int_0^1 \frac{10x^2+24x+8}{2x^3+11x^2+6x+5}~\dee{x}&= \int_0^1 \left( \frac{3}{x+5} + \frac{4x+1}{2x^2+x+1}\right) ~\dee{x}\end{align*}

    The antiderivative of the left fraction is 3log⁡∣x+5∣3\log|x+5|. For the right fraction, we use the substitution u=2x2+x+1u=2x^2+x+1, du=(4x+1)dx\dee{u}=(4x+1)\dee{x} to antidifferentiate.

    =[3log⁡∣x+5∣+log⁡∣2x2+x+1∣]01=3log⁡6+log⁡4−3log⁡5−log⁡1=log⁡(4⋅6353)\begin{align*}&=\big[3\log|x+5| + \log|2x^2+x+1|\big]_0^1\\ &=3\log 6 + \log 4 - 3\log 5 -\log 1\\ &=\log \left(\frac{4\cdot 6^3}{5^3}\right)\end{align*}

3Stage 3Application

Further than practice: several ideas at once, or an unfamiliar situation.

In Questions 18 and 19, we use partial fraction to find the antiderivatives of two important functions: cosecant, and cosecant cubed.

Q18Stage 3

Using the method of Example 1.10.5 in the CLP-2 text, integrate ∫csc⁡x dx\displaystyle\int \csc x ~\dee{x}.

Hint

csc⁡x=1sin⁡x=sin⁡xsin⁡2x\displaystyle\csc x = \frac{1}{\sin x} = \frac{\sin x}{\sin^2 x}

Answer

12log⁡∣1−cos⁡x1+cos⁡x∣+C\displaystyle\frac{1}{2}\log\left| \frac{1-\cos x}{1+\cos x}\right|+C

Full solution

We follow the example in the text.

∫csc⁡x dx=∫1sin⁡x dx=∫sin⁡xsin⁡2x dx=∫sin⁡x1−cos⁡2x dx\begin{align*}\int \csc x ~\dee{x}&=\int \frac{1}{\sin x} ~\dee{x}=\int \frac{\sin x}{\sin^2 x} ~\dee{x} = \int \frac{\sin x}{1-\cos^2 x} ~\dee{x}\end{align*}

Let u=cos⁡xu=\cos x, du=−sin⁡x dx\dee{u}=-\sin x~\dee{x}.

=∫−11−u2 du=∫−1(1+u)(1−u) du\begin{align*}&=\int\frac{-1}{1-u^2}~\dee{u} = \int\frac{-1}{(1+u)(1-u)}~\dee{u}\end{align*}

We see an opportunity for partial fraction.

−1(1+u)(1−u)=A1+u+B1−u\begin{align*}\frac{-1}{(1+u)(1-u)}&=\frac{A}{1+u}+\frac{B}{1-u}\end{align*}

Multiply both sides by the original denominator.

−1=A(1−u)+B(1+u)\begin{align*}-1&=A(1-u)+B(1+u)\end{align*}

Let u=1u=1.

−1=2B⇒B=−12\begin{align*}-1&=2B \qquad\Rightarrow\color{red} B = -\frac{1}{2}\end{align*}

Let u=−1u=-1.

−1=2A⇒A=−12\begin{align*}-1&=2A \qquad\Rightarrow \color{red}A = -\frac{1}{2}\end{align*}

We can now re-write our integral.

∫csc⁡x dx=∫−1(1+u)(1−u) du=∫(−1/21+u+−1/21−u) du=−12log⁡∣1+u∣+12log⁡∣1−u∣+C=12log⁡∣1−u1+u∣+C=12log⁡∣1−cos⁡x1+cos⁡x∣+C\begin{align*}\int \csc x ~\dee{x}&= \int\frac{-1}{(1+u)(1-u)}~\dee{u} =\int \left(\frac{-1/2}{1+u} + \frac{-1/2}{1-u}\right)~\dee{u}\\ &=-\frac{1}{2}\log|1+u| +\frac{1}{2}\log|1-u|+C\\ &=\frac{1}{2}\log\left| \frac{1-u}{1+u}\right|+C\\ &=\frac{1}{2}\log\left| \frac{1-\cos x}{1+\cos x}\right|+C\end{align*}

Remark: Elsewhere in the text, and in many tables of integrals, the antiderivative of cosecant is given as log⁡∣csc⁡x−cot⁡x∣\log|\csc x - \cot x|. We show that this is equivalent to our result.

log⁡∣csc⁡x−cot⁡x∣=12log⁡∣(csc⁡x−cot⁡x)2∣=12log⁡∣csc⁡2x−2csc⁡xcot⁡x+cot⁡2x∣=12log⁡∣1sin⁡2x−2cos⁡xsin⁡2x+cos⁡2xsin⁡2x∣=12log⁡∣1−2cos⁡x+cos⁡2xsin⁡2x∣=12log⁡∣(1−cos⁡x)21−cos⁡2x∣=12log⁡∣(1−cos⁡x)2(1−cos⁡x)(1+cos⁡x)∣=12log⁡∣1−cos⁡x1+cos⁡x∣\begin{align*} \log|\csc x - \cot x|&= \frac{1}{2}\log\left|\left(\csc x - \cot x\right)^2\right| = \frac{1}{2}\log\left| \csc^2 x - 2\csc x \cot x + \cot^2 x\right|\\ &=\frac{1}{2}\log\left| \frac{1}{\sin^2x} - \frac{2\cos x}{\sin^2 x}+\frac{\cos^2 x}{\sin^2x}\right|\\ &=\frac{1}{2}\log\left| \frac{1-2\cos x + \cos^2 x}{\sin^2x} \right| =\frac{1}{2}\log\left| \frac{\left(1-\cos x\right)^2}{1-\cos^2x} \right| \\&=\frac{1}{2}\log\left| \frac{\left(1-\cos x\right)^2}{(1-\cos x)(1+\cos x)} \right| =\frac{1}{2}\log\left| \frac{1-\cos x}{1+\cos x} \right| \end{align*}
Q19Stage 3

Using the method of Example 1.10.6 in the CLP-2 text, integrate ∫csc⁡3x dx\displaystyle\int \csc^3 x ~\dee{x}.

Hint

Use the partial fraction decomposition from Queston 18 to save yourself some time.

Answer

−cos⁡x2sin⁡2x+14log⁡∣1−cos⁡x1+cos⁡x∣+C\displaystyle\frac{-\cos x}{2\sin^2 x} + \frac{1}{4}\log\left|\frac{1-\cos x}{1+\cos x}\right|+C

Full solution

We follow the example in the text.

∫csc⁡3x dx=∫1sin⁡3x dx=∫sin⁡xsin⁡4x dx=∫sin⁡x(1−cos⁡2x)2 dx\begin{align*}\displaystyle\int \csc^3 x ~\dee{x}&=\int \frac{1}{\sin^3x}~\dee{x} =\int \frac{\sin x}{\sin^4x}~\dee{x}=\int \frac{\sin x}{(1-\cos^2x)^2}~\dee{x}\end{align*}

Let u=cos⁡xu=\cos x, du=−sin⁡x dx\dee{u}=-\sin x~\dee{x}.

=∫−1(1−u2)2 du\begin{align*}&=\int\frac{-1}{(1-u^2)^2}~\dee{u}\end{align*}

In Question 18, we saw 11−u2=1/21+u+1/21−u\frac{1}{1-u^2} = \frac{1/2}{1+u}+\frac{1/2}{1-u} , so

∫−1(1−u2)2 du=−∫(11−u2)2 du=−∫(1/21+u+1/21−u)2 du=−14∫(1(1+u)2+21−u2+1(1−u)2) du=−14∫(1(1+u)2+11+u+11−u+1(1−u)2) du=−14(−11+u+log⁡∣1+u∣−log⁡∣1−u∣+11−u)+C=−14(2u1−u2+log⁡∣1+u1−u∣)+C=−u2(1−u2)+14log⁡∣1−u1+u∣+C=−cos⁡x2sin⁡2x+14log⁡∣1−cos⁡x1+cos⁡x∣+C\begin{align*}\int\frac{-1}{(1-u^2)^2}~\dee{u}&=-\int\left(\textcolor{blue}{\frac{1}{1-u^2}}\right)^2~\dee{u} =-\int\left(\textcolor{blue}{\frac{1/2}{1+u}+\frac{1/2}{1-u}}\right)^2~\dee{u}\\ &=-\frac{1}{4}\int\left(\frac{1}{(1+u)^2}+\textcolor{blue}{\frac{2}{1-u^2}}+\frac{1}{(1-u)^2}\right)~\dee{u} \\ &=-\frac{1}{4}\int\left(\frac{1}{(1+u)^2}+ \textcolor{blue}{\frac{1}{1+u}+\frac{1}{1-u}}+\frac{1}{(1-u)^2}\right)~\dee{u}\\ &=-\frac{1}{4}\left(-\frac{1}{1+u} + \log|1+u|-\log|1-u|+\frac{1}{1-u}\right)+C\\ &=-\frac{1}{4}\left(\frac{2u}{1-u^2} + \log\left|\frac{1+u}{1-u}\right|\right)+C\\ &=\frac{-u}{2(1-u^2)} + \frac{1}{4}\log\left|\frac{1-u}{1+u}\right|+C\\ &=\frac{-\cos x}{2\sin^2 x} + \frac{1}{4}\log\left|\frac{1-\cos x}{1+\cos x}\right|+C\end{align*}

Remark: In Example 1.8.23 of the CLP-2 text, and in many tables of integrals, the antiderivative of csc⁡3x\csc^3 x is given as −12cot⁡xcsc⁡x+12log⁡∣csc⁡x−cot⁡x∣+C-\frac{1}{2}\cot x \csc x + \frac{1}{2}\log|\csc x - \cot x|+C. This is equivalent to our result. Recall in the remark after the solution to Question 18, we saw 12log⁡∣1−cos⁡x1+cos⁡x∣=log⁡∣csc⁡x−cot⁡x∣\frac{1}{2}\log\left| \frac{1-\cos x}{1+\cos x}\right|=\log|\csc x - \cot x|.

−12cot⁡xcsc⁡x+12log⁡∣csc⁡x−cot⁡x∣=−12cot⁡xcsc⁡x+14log⁡∣1−cos⁡x1+cos⁡x∣=−12(cos⁡xsin⁡x)(1sin⁡x)+14log⁡∣1−cos⁡x1+cos⁡x∣=−cos⁡x2sin⁡2x+14log⁡∣1−cos⁡x1+cos⁡x∣\begin{align*} -\frac{1}{2}\cot x \csc x + \frac{1}{2}\log|\csc x - \cot x|&= -\frac{1}{2}\cot x \csc x + \frac{1}{4}\log\left| \frac{1-\cos x}{1+\cos x}\right|\\ &=-\frac{1}{2}\left(\frac{\cos x}{\sin x}\right)\left(\frac{1}{\sin x}\right) + \frac{1}{4}\log\left| \frac{1-\cos x}{1+\cos x}\right|\\ &=\frac{-\cos x}{2\sin^2 x }+ \frac{1}{4}\log\left| \frac{1-\cos x}{1+\cos x}\right| \end{align*}

The purpose of performing a partial fraction decomposition is to manipulate an integrand into a form that is easily integrable. These “easily integrable" forms are rational functions whose denominator is a power of a linear function, or of an irreducible quadratic function. In Questions 20 through 23, we explore the integration of rational functions whose denominators involve irreducible quadratics.

Q20Stage 3

Evaluate ∫123x3+15x2+35x+10x4+5x3+10x2 dx\displaystyle\int_1^2 \frac{3x^3+15x^2+35x+10}{x^4+5x^3+10x^2}~\dee{x}.

Hint

In the final integration, complete the square to make a piece of the integrand look more like the derivative of arctangent.

Answer

3log⁡2+12+215(arctan⁡(715)−arctan⁡(915))\displaystyle3\log 2 + \frac{1}{2}+\frac{2}{\sqrt{15}}\left(\arctan\left(\frac{7}{\sqrt{15}}\right)-\arctan\left(\frac{9}{\sqrt{15}}\right)\right)

Full solution

This is a rational function, and there's no obvious substitution, so we'll use partial fraction decomposition.

  • First, we check that the numerator has strictly smaller degree than the denominator, so we don't have to use long division.

  • Second, we factor the denominator. We can immediately pull out a factor of x2x^2; then we're left with the quadratic polynomial x2+5x+10x^2+5x+10. Using the quadratic equation, we check that this has no real roots, so it is irreducible.

  • Once we know the factorization of the denominator, we can set up our decomposition.

    3x3+15x2+35x+10x4+5x3+10x2=3x3+15x2+35x+10x2(x2+5x+10)=Ax+Bx2+Cx+Dx2+5x+10\begin{align*}\frac{3x^3+15x^2+35x+10}{x^4+5x^3+10x^2}&= \frac{3x^3+15x^2+35x+10}{x^2(x^2+5x+10)}\\ &=\frac{A}{x}+\frac{B}{x^2}+\frac{Cx+D}{x^2+5x+10}\end{align*}

    We multiply both sides by the original denominator.

    3x3+15x2+35x+10=Ax(x2+5x+10)+B(x2+5x+10)+(Cx+D)x2\begin{align*}3x^3+15x^2+35x+10&=Ax(x^2+5x+10) + B(x^2+5x+10)+(Cx+D)x^2 \tag{1}\end{align*}

    Following the “Sneaky Method," we plug in x=0x=0.

    0+10=A(0)+B(10)+(C(0)+D)(0)B=1\begin{align*}0+10&=A(0)+B(10)+(C(0)+D)(0)\\ \color{red}B&\color{red}=1\end{align*}

  • Knowing BB allows us to simplify our Equation (1).

    3x3+15x2+35x+10=Ax(x2+5x+10)+1(x2+5x+10)+(Cx+D)x23x3+14x2+30x=Ax(x2+5x+10)+(Cx+D)x2\begin{align*}3x^3+15x^2+35x+10&=Ax(x^2+5x+10) + \textcolor{red}{1}(x^2+5x+10)+(Cx+D)x^2\\ 3x^3+14x^2+30x&=Ax(x^2+5x+10) +(Cx+D)x^2\end{align*}

    We can factor xx out of both sides of the equation.

    3x2+14x+30=A(x2+5x+10)+(Cx+D)x\begin{align*}3x^2+14x+30&=A(x^2+5x+10) +(Cx+D)x \tag{2}\end{align*}

  • Again, we set x=0x=0.

    0+30=A(10)+(C(0+D)(0)A=3\begin{align*} 0+30&=A(10) +(C(0+D)(0)\\ \color{red}A&\color{red}=3 \end{align*}

  • We simplify Equation (2), using A=3A=3.

    3x2+14x+30=3(x2+5x+10)+(Cx+D)x−x=Cx2+DxC=0,D=−1\begin{align*} 3x^2+14x+30&=\textcolor{red}{3}(x^2+5x+10) +(Cx+D)x\\ -x&=Cx^2+Dx\\ \color{red}C&\color{red}=0,\quad D=-1 \end{align*}

  • Now that we have our coefficients, we can re-write our integral in a friendlier form.

    ∫123x3+15x2+35x+10x4+5x+10x2 dx=∫12(3x+1x2−1x2+5x+10) dx=[3log⁡∣x∣−1x]12−∫121x2+5x+10 dx=3log⁡2+12−∫121x2+5x+10 dx\begin{align*}\int_1^2 \frac{3x^3+15x^2+35x+10}{x^4+5x+10x^2}~\dee{x}&= \int_1^2 \left(\frac{3}{x}+\frac{1}{x^2}-\frac{1}{x^2+5x+10} \right)~\dee{x}\\ &=\left[3\log|x|-\frac{1}{x}\right]_1^2-\int_1^2\frac{1}{x^2+5x+10} ~\dee{x}\\ &=3\log 2+\frac{1}{2}-\int_1^2\frac{1}{x^2+5x+10} ~\dee{x}\end{align*}

    The remaining integral is the reciprocal of a quadratic polynomial, much like 11+x2\dfrac{1}{1+x^2}, whose antiderivative is arctangent. We complete the square and use the substitution u=(2x+515)u=\left(\frac{2x+5}{\sqrt{15}}\right), du=215 dx\dee{u}=\frac{2}{\sqrt{15}}~\dee{x}.

    ∫121x2+5x+10 dx=∫121(x+52)2+154 dx=415∫121(2x+515)2+1 dx=215∫7/159/151u2+1 du=215[arctan⁡u]7/159/15=215(arctan⁡(915)−arctan⁡(715))\begin{align*}\int_1^2 \frac{1}{x^2+5x+10}~\dee{x}&=\int_1^2 \frac{1}{\left(x+\frac{5}{2}\right)^2+\frac{15}{4}}~\dee{x}\\ &=\frac{4}{15}\int_1^2 \frac{1}{\left(\frac{2x+5}{\sqrt{15}}\right)^2+1}~\dee{x} \\&=\frac{2}{\sqrt{15}}\int_{7/\sqrt{15}}^{9/\sqrt{15}} \frac{1}{u^2+1}~\dee{u} \\&=\frac{2}{\sqrt{15}}\Big[\arctan u\Big]_{7/\sqrt{15}}^{9/\sqrt{15}} \\&=\frac{2}{\sqrt{15}}\left(\arctan\left(\frac{9}{\sqrt{15}}\right)-\arctan\left(\frac{7}{\sqrt{15}}\right)\right)\end{align*}

    So, all together,

    ∫123x3+15x2+35x+10x4+5x3+10x2 dx=3log⁡2+12−215(arctan⁡(915)−arctan⁡(715))\int_1^2 \frac{3x^3+15x^2+35x+10}{x^4+5x^3+10x^2}~\dee{x}=3\log 2 + \frac{1}{2}-\frac{2}{\sqrt{15}}\left(\arctan\left(\frac{9}{\sqrt{15}}\right)-\arctan\left(\frac{7}{\sqrt{15}}\right)\right)
Q21Stage 3

Evaluate ∫(3x2+2+x−3(x2+2)2) dx\displaystyle\int\left(\frac{3}{x^2+2}+\frac{x-3}{(x^2+2)^2}\right) ~\dee{x}.

Hint

Review Question 20 in Section 1.9 for antidifferentiation tips.

Answer

=942arctan⁡(x2)−2+3x4(x2+2)+C\displaystyle=\frac{9}{4\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)-\frac{2+3x}{4(x^2+2)} +C

Full solution

Our integrand is already in the nice form that would come out of a partial fractions decomposition. Let's consider its different pieces.

  • First piece: ∫3x2+2 dx\int \frac{3}{x^2+2}~\dee{x}. The fraction looks somewhat like the derivative of arctangent, so we can massage it to find an appropriate substitution.

    ∫3x2+2 dx=32∫1(x2)2+1 dx\begin{align*}\int \frac{3}{x^2+2}~\dee{x}&=\frac{3}{2}\int \frac{1}{\left(\frac{x}{\sqrt2}\right)^2+1}~\dee{x}\end{align*}

    Use the substitution u=x2u=\frac{x}{\sqrt2}, du=12 dx.\dee{u} = \frac{1}{\sqrt{2}}~\dee{x}.

    =32∫1u2+1 du=32arctan⁡u+C=32arctan⁡(x2)+C\begin{align*}&=\frac{3}{\sqrt2}\int\frac{1}{u^2+1}~\dee{u}\\ &=\frac{3}{\sqrt2}\arctan u+C\\ &=\frac{3}{\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)+C\end{align*}

  • The next piece is ∫x−3(x2+2)2 dx\int \frac{x-3}{(x^2+2)^2}~\dee{x}. If the numerator were only xx (and no constant), we could use the substitution u=x2+2u=x^2+2, du=2x dx\dee{u} = 2x~\dee{x}. So, to that end, we can break up that fraction into x(x2+2)2−3(x2+2)2\frac{x}{(x^2+2)^2}-\frac{3}{(x^2+2)^2}. For now, we only evaluate the first half.

    ∫x(x2+2)2 dx=12∫1u2 du=−12u+C=−12x2+4+C\begin{align*} \int \frac{x}{(x^2+2)^2}~\dee{x}&=\frac{1}{2}\int\frac{1}{u^2}~\dee{u} = -\frac{1}{2u}+C\\ &=-\frac{1}{2x^2+4}+C \end{align*}

  • That leaves us with the final piece, 3(x2+2)2\frac{3}{(x^2+2)^2}, which is the hardest. We saw something similar in Question 20 in Section 1.9: we can use the substitution x=2tan⁡θx=\sqrt{2}\tan\theta, dx=2sec⁡2θ dθ\dee{x} = \sqrt{2}\sec^2\theta~\dee{\theta}.

    ∫3(x2+2)2 dx=∫3(2tan⁡2θ+2)22sec⁡2θ dθ=∫34sec⁡4θ2sec⁡2θ dθ=322∫cos⁡2θ dθ=342∫(1+cos⁡(2θ)) dθ=342(θ+12sin⁡(2θ))+C=342(θ+sin⁡θcos⁡θ)+C=342(arctan⁡(x2)+x2x2+2)+C\begin{align*} \int\frac{3}{(x^2+2)^2}~\dee{x}&=\int\frac{3}{(2\tan^2\theta+2)^2}\sqrt{2}\sec^2\theta~\dee{\theta}\\ &=\int\frac{3}{4\sec^4\theta}\sqrt{2}\sec^2\theta~\dee{\theta}\\ &=\frac{3}{2\sqrt{2}}\int\cos^2\theta~\dee{\theta}\\ &=\frac{3}{4\sqrt{2}}\int\big(1+\cos(2\theta)\big)~\dee{\theta}\\ &=\frac{3}{4\sqrt{2}}\left(\theta+\frac{1}{2}\sin(2\theta)\right)+C\\ &=\frac{3}{4\sqrt{2}}\left(\theta+\sin\theta\cos\theta\right)+C\\ \hspace{2cm} &=\frac{3}{4\sqrt{2}}\left(\arctan\left(\frac{x}{\sqrt2}\right)+\frac{x\sqrt{2}}{x^2+2}\right)+C \end{align*}

    Figure from prob_s1.10, line 1168

    Figure from prob_s1.10, line 1168

    From our substitution, tan⁡θ=x2\tan\theta = \frac{x}{\sqrt2}. So, we can draw a right triangle with angle θ\theta, opposite side xx, and adjacent side 2\sqrt2. Then by the Pythagorean Theorem, the hypotenuse has length x2+2\sqrt{x^2+2}, and this gives us sin⁡θ\sin\theta and cos⁡θ\cos\theta.

Now we have our integral.

∫(3x2+2+x−3(x2+2)2) dx=∫3x2+2 dx+∫x(x2+2)2 dx−∫3(x2+2)2 dx=32arctan⁡(x2)−12x2+4−342(arctan⁡(x2)+x2x2+2)+C=942arctan⁡(x2)−12(x2+2)−3x4(x2+2)+C=942arctan⁡(x2)−2+3x4(x2+2)+C\begin{align*} \int\bigg(\frac{3}{x^2+2}&+\frac{x-3}{(x^2+2)^2}\bigg) ~\dee{x}= \int\frac{3}{x^2+2}~\dee{x}+\int\frac{x}{(x^2+2)^2} ~\dee{x}-\int\frac{3}{(x^2+2)^2} ~\dee{x}\\ &=\frac{3}{\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)-\frac{1}{2x^2+4} -\frac{3}{4\sqrt{2}}\left(\arctan\left(\frac{x}{\sqrt2}\right)+\frac{x\sqrt{2}}{x^2+2}\right)+C \\ &=\frac{9}{4\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)-\frac{1}{2(x^2+2)} -\frac{3x}{4(x^2+2)}+C \\ &=\frac{9}{4\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)-\frac{2+3x}{4(x^2+2)} +C \end{align*}
Q22Stage 3

Evaluate ∫1(1+x2)3 dx\displaystyle\int\frac{1}{(1+x^2)^3}~\dee{x}.

Hint

Partial fraction decomposition won't simplify this any more. Use a trig substitution.

Answer

38arctan⁡x+3x3+5x8(1+x2)2+C\displaystyle\frac{3}{8}\arctan x + \frac{3x^3+5x}{8(1+x^2)^2}+C

Full solution

This is already as simplified as we can make it using partial fraction. Indeed, this is the kind of term that could likely come out of the partial fraction decomposition of a scarier rational function. So, we need to know how to integrate it. Similar to the last piece we integrated in Question 21, we can use the substitution x=tan⁡θx=\tan\theta, dx=sec⁡2θ dθ.\dee{x} = \sec^2\theta~\dee{\theta}.

∫1(1+x2)3 dx=∫sec⁡2θ(1+tan⁡2θ)3 dθ=∫sec⁡2θ(sec⁡2θ)3 dθ=∫cos⁡4θ dθ=∫[1+cos⁡(2θ)2]2 dθ=14∫(1+cos⁡(2θ))2 dθ=14∫(1+2cos⁡(2θ)+cos⁡2(2θ)) dθ=14∫(1+2cos⁡(2θ)+12(1+cos⁡(4θ))) dθ=14∫(32+2cos⁡(2θ)+12cos⁡(4θ))) dθ=14(32θ+sin⁡(2θ)+18sin⁡(4θ))+C=38θ+14sin⁡(2θ)+132sin⁡(4θ)+C=38θ+12sin⁡θcos⁡θ+116sin⁡(2θ)cos⁡(2θ)+C=38θ+12sin⁡θcos⁡θ+18sin⁡θcos⁡θ(cos⁡2θ−sin⁡2θ)+C=38arctan⁡x+x2(1+x2)+18(x1+x2)(1−x21+x2)+C=38arctan⁡x+3x3+5x8(1+x2)2+C\begin{align*} \int\frac{1}{(1+x^2)^3}~\dee{x}&=\int \frac{\sec^2\theta}{(1+\tan^2\theta)^3}~\dee{\theta} =\int \frac{\sec^2\theta}{(\sec^2\theta)^3}~\dee{\theta}\\ &=\int \cos^4\theta~\dee{\theta} =\int \left[\frac{1+\cos(2\theta)}{2}\right]^2~\dee{\theta} \\ &= \frac{1}{4}\int (1+\cos(2\theta))^2~\dee{\theta}\\ &=\frac{1}{4}\int\left(1+2\cos(2\theta)+\cos^2(2\theta)\right)~\dee{\theta}\\ &=\frac{1}{4}\int\left(1+2\cos(2\theta)+\frac{1}{2}(1+\cos(4\theta))\right)~\dee{\theta}\\ &=\frac{1}{4}\int\left(\frac{3}{2}+2\cos(2\theta)+\frac{1}{2}\cos(4\theta))\right)~\dee{\theta}\\ &=\frac{1}{4}\left(\frac{3}{2}\theta + \sin(2\theta) +\frac{1}{8}\sin(4\theta)\right)+C\\ &=\frac{3}{8}\theta + \frac{1}{4}\sin(2\theta) +\frac{1}{32}\sin(4\theta)+C\\ &=\frac{3}{8}\theta + \frac{1}{2}\sin\theta\cos\theta +\frac{1}{16}\sin(2\theta)\cos(2\theta)+C\\ &=\frac{3}{8}\theta + \frac{1}{2}\sin\theta\cos\theta +\frac{1}{8}\sin\theta\cos\theta(\cos^2\theta-\sin^2\theta)+C\\ &=\frac{3}{8}\arctan x + \frac{x}{2(1+x^2)}+\frac{1}{8}\left(\frac{x}{1+x^2}\right)\left(\frac{1-x^2}{1+x^2}\right)+C\\ \hspace{2cm}&=\frac{3}{8}\arctan x + \frac{3x^3+5x}{8(1+x^2)^2}+C \end{align*}

Figure from prob_s1.10, line 1234

Figure from prob_s1.10, line 1234

To change our variables from θ\theta to xx, recall we used the substitution x=tan⁡θx=\tan\theta. So, we draw a right triangle with angle θ\theta, opposite side length xx, and adjacent side length 1. By the Pythagorean Theorem, the hypotenuse has length 1+x2\sqrt{1+x^2}. This allows us to find sin⁡θ\sin\theta and cos⁡θ\cos\theta.

Q23Stage 3

Evaluate ∫(3x+3x+1x2+5+3x(x2+5)2) dx\displaystyle\int \left(3x+\frac{3x+1}{x^2+5}+\frac{3x}{(x^2+5)^2}\right) ~\dee{x}.

Hint

To evaluate the antiderivative, break one of the fractions into two fractions.

Answer

32x2+15arctan⁡(x5)+32log⁡∣x2+5∣−32x2+10+C\displaystyle\frac{3}{2}x^2+\frac{1}{\sqrt{5}}\arctan \left(\frac{x}{\sqrt5}\right)+ \frac{3}{2}\log|x^2+5|-\frac{3}{2x^2+10}+C

Full solution

Our integrand is already as simplified as the method of partial fractions can make it. The first term is easy to antidifferentiate. The second term would be easier if it were broken into two pieces: one where the numerator is a constant, and one where the numerator is a multiple of xx.

∫(3x+3x+1x2+5+3x(x2+5)2) dx=32x2+∫(1x2+5+3xx2+5+3x(x2+5)2) dx=32x2+∫1x2+5dx+∫(3xx2+5+3x(x2+5)2) dx\begin{align*}\int \left(3x+\frac{3x+1}{x^2+5}+\frac{3x}{(x^2+5)^2}\right) ~\dee{x} &= \frac{3}{2}x^2+\int\left(\frac{1}{x^2+5}+\frac{3x}{x^2+5}+\frac{3x}{(x^2+5)^2}\right) ~\dee{x}\\ &= \frac{3}{2}x^2+\int\frac{1}{x^2+5}\dee{x}+\int\left(\frac{3x}{x^2+5}+\frac{3x}{(x^2+5)^2}\right) ~\dee{x}\end{align*}

The first integral looks similar to the derivative of arctangent. For the second integral, we use the substitution u=x2+5u=x^2+5, du=2x dx\dee{u}=2x~\dee{x}.

=32x2+15∫1(x5)2+1dx+∫(3/2u+3/2u2) du\begin{align*}&= \frac{3}{2}x^2+\frac{1}{5}\int\frac{1}{\left(\frac{x}{\sqrt5}\right)^2+1}\dee{x}+\int\left(\frac{3/2}{u}+\frac{3/2}{u^2}\right) ~\dee{u}\end{align*}

For the first integral, use the substitution w=x5w=\frac{x}{\sqrt5}, dw=15 dx\dee{w} = \frac{1}{\sqrt5}~\dee{x}.

=32x2+15∫1w2+1dw+32log⁡∣u∣−32u=32x2+15arctan⁡w+32log⁡∣x2+5∣−32x2+10+C=32x2+15arctan⁡(x5)+32log⁡∣x2+5∣−32x2+10+C\begin{align*}&= \frac{3}{2}x^2+\frac{1}{\sqrt{5}}\int\frac{1}{w^2+1}\dee{w}+ \frac{3}{2}\log|u|-\frac{3}{2u}\\ &= \frac{3}{2}x^2+\frac{1}{\sqrt{5}}\arctan w+ \frac{3}{2}\log|x^2+5|-\frac{3}{2x^2+10}+C\\ &= \frac{3}{2}x^2+\frac{1}{\sqrt{5}}\arctan \left(\frac{x}{\sqrt5}\right)+ \frac{3}{2}\log|x^2+5|-\frac{3}{2x^2+10}+C\end{align*}

In Questions 24 through 26, we use substitution to turn a non-rational integrand into a rational integrand, then evaluate the resulting integral using partial fraction. Till now, the partial fraction problems you've seen have all looked largely the same, but keep in mind that a partial fraction decomposition can be a small step in a larger problem.

Q24Stage 3

Evaluate ∫cos⁡θ3sin⁡θ+cos⁡2θ−3 dθ\displaystyle\int \frac{\cos\theta}{3\sin\theta+\cos^2\theta-3} ~\dee{\theta}.

Hint

cos⁡2θ=1−sin⁡2θ\cos^2\theta = 1-\sin^2\theta

Answer

log⁡∣sin⁡θ−1sin⁡θ−2∣+C\displaystyle\log\left| \frac{\sin\theta-1}{\sin\theta-2}\right|+C

Full solution

If our denominator were all sines, we could use the substitution x=sin⁡θx=\sin\theta. To that end, we apply the identity cos⁡2θ=1−sin⁡2θ\cos^2\theta =1- \sin^2\theta.

∫cos⁡θ3sin⁡θ+cos⁡2θ−3 dθ=∫cos⁡θ3sin⁡θ+1−sin⁡2θ−3 dθ=∫cos⁡θ3sin⁡θ−sin⁡2θ−2 dθ\begin{align*}\int \frac{\cos\theta}{3\sin\theta+\cos^2\theta-3} ~\dee{\theta}&= \int \frac{\cos\theta}{3\sin\theta+1-\sin^2\theta-3} ~\dee{\theta}= \int \frac{\cos\theta}{3\sin\theta-\sin^2\theta-2} ~\dee{\theta}\end{align*}

We use the substitution x=sin⁡θx=\sin \theta, dx=cos⁡θ dθ\dee{x}=\cos\theta~\dee{\theta}.

=∫13x−x2−2 dx=∫−1x2−3x+2 dx=∫−1(x−1)(x−2) dx\begin{align*}&=\int \frac{1}{3x-x^2-2}~\dee{x}=\int \frac{-1}{x^2-3x+2}~\dee{x} =\int \frac{-1}{(x-1)(x-2)}~\dee{x}\end{align*}

Now we can find a partial fraction decomposition.

−1(x−1)(x−2)=Ax−1+Bx−2−1=A(x−2)+B(x−1)\begin{align*}\frac{-1}{(x-1)(x-2)}&=\frac{A}{x-1}+\frac{B}{x-2}\\ -1&=A(x-2)+B(x-1)\end{align*}

Setting x=1x=1 and x=2x=2, we see

A=1,B=−1\begin{align*}\color{red}A&\color{red}=1\textcolor{black}{,}\quad B=-1\end{align*}

Now, we can evaluate our integral.

∫cos⁡θ3sin⁡θ+cos⁡2θ−3 dθ=∫−1(x−1)(x−2) dx=∫(1x−1−1x−2) dx=log⁡∣x−1∣−log⁡∣x−2∣+C=log⁡∣x−1x−2∣+C=log⁡∣sin⁡θ−1sin⁡θ−2∣+C\begin{align*}\int \frac{\cos\theta}{3\sin\theta+\cos^2\theta-3} ~\dee{\theta}&=\int \frac{-1}{(x-1)(x-2)}~\dee{x} =\int \left(\frac{1}{x-1} - \frac{1}{x-2}\right)~\dee{x}\\ &=\log|x-1| - \log|x-2|+C=\log\left| \frac{x-1}{x-2}\right|+C\\ &=\log\left| \frac{\sin\theta-1}{\sin\theta-2}\right|+C\end{align*}
Q25Stage 3

Evaluate ∫1e2t+et+1 dt\displaystyle\int\frac{1}{e^{2t}+e^t+1}~\dee{t}.

Hint

If you're having a hard time making the substitution, multiply the numerator and the denominator by exe^x.

Answer

t−12log⁡∣e2t+et+1∣−13arctan⁡(2et+13)+C\displaystyle t - \frac{1}{2}\log|e^{2t}+e^t+1|-\frac{1}{\sqrt3}\arctan \left(\frac{2e^t+1}{\sqrt3}\right)+C

Full solution

This looks a lot like a rational function, but with the function ete^t in place of the variable. So, we would like to make the substitution x=etx=e^t, dx=etdt\dee{x}=e^t\dee{t}. Then dt=1et dx=1x dx\dee{t} = \frac{1}{e^t}\,\dee{x} = \frac{1}{x}\,\dee{x}.

∫1e2t+et+1 dt=∫1x(x2+x+1)dx\begin{align*}\int\frac{1}{e^{2t}+e^t+1}~\dee{t}&=\int\frac{1}{x\left(x^2+x+1\right)}\dee{x}\end{align*}

The factor x2+x+1x^2+x+1 is an irreducible quadratic, so the denominator is completely factored. Now we can use partial fraction decomposition.

1x(x2+x+1)=Ax+Bx+Cx2+x+11=A(x2+x+1)+(Bx+C)x1=(A+B)x2+(A+C)x+A\begin{align*}\frac{1}{x\left(x^2+x+1\right)}&=\frac{A}{x}+\frac{Bx+C}{x^2+x+1}\\ 1&=A(x^2+x+1)+(Bx+C)x\\ 1&=(A+B)x^2+(A+C)x+A\end{align*}

The constant terms tell us A=1A=1; then the coefficient of xx tells us C=−A=−1\textcolor{red}{C=-A=-1}. Finally, the coefficient of x2x^2 tells us B=−A=−1\textcolor{red}{B=-A=-1}. Now we can evaluate our integral.

∫1e2t+et+1 dt=∫1x(x2+x+1)dx=∫(1x−x+1x2+x+1) dx=∫(1x−x+1/2+1/2x2+x+1) dx=∫1x dx−∫x+1/2x2+x+1 dx−∫1/2x2+x+1 dx=log⁡∣x∣−12log⁡∣x2+x+1∣−∫1/2x2+x+1 dx\begin{align*}\int\frac{1}{e^{2t}+e^t+1}~\dee{t}&=\int\frac{1}{x\left(x^2+x+1\right)}\dee{x}\\ &=\int\left(\frac{1}{x} - \frac{x+1}{x^2+x+1}\right)~\dee{x}\\ &=\int\left(\frac{1}{x} - \frac{x+1/2+1/2}{x^2+x+1}\right)~\dee{x}\tag{$*$}\\ &=\int\frac{1}{x} ~\dee{x}-\int \frac{x+1/2}{x^2+x+1}~\dee{x}-\int \frac{1/2}{x^2+x+1}~\dee{x}\\ &=\log|x| - \frac{1}{2}\log|x^2+x+1|-\int \frac{1/2}{x^2+x+1}~\dee{x}\end{align*}

In step (∗*), we set ourselves up so that we could evaluate the second integral with the substitution u=x2+x+1u=x^2+x+1. For the remaining integral, we complete the square, so that the integrand looks something like the derivative of arctangent.

=log⁡∣x∣−12log⁡∣x2+x+1∣−∫1/2(x+12)2+34 dx=log⁡∣x∣−12log⁡∣x2+x+1∣−23∫1(2x+13)2+1 dx\begin{align*}&=\log|x| - \frac{1}{2}\log|x^2+x+1|-\int \frac{1/2}{\left(x+\frac{1}{2}\right)^2+\frac{3}{4}}~\dee{x} \\&=\log|x| - \frac{1}{2}\log|x^2+x+1|-\frac{2}{3}\int \frac{1}{\left(\frac{2x+1}{\sqrt3}\right)^2+1}~\dee{x}\end{align*}

We use the substitution u=2x+13u=\frac{2x+1}{\sqrt{3}}, du=23\dee{u} = \frac{2}{\sqrt3}.

=log⁡∣x∣−12log⁡∣x2+x+1∣−13∫1u2+1 du=log⁡∣x∣−12log⁡∣x2+x+1∣−13arctan⁡u+C=log⁡∣x∣−12log⁡∣x2+x+1∣−13arctan⁡(2x+13)+C=log⁡∣et∣−12log⁡∣e2t+et+1∣−13arctan⁡(2et+13)+C=t−12log⁡∣e2t+et+1∣−13arctan⁡(2et+13)+C\begin{align*}&=\log|x| - \frac{1}{2}\log|x^2+x+1|-\frac{1}{\sqrt3}\int \frac{1}{u^2+1}~\dee{u} \\&=\log|x| - \frac{1}{2}\log|x^2+x+1|-\frac{1}{\sqrt3}\arctan u+C \\&=\log|x| - \frac{1}{2}\log|x^2+x+1|-\frac{1}{\sqrt3}\arctan \left(\frac{2x+1}{\sqrt3}\right)+C \\&=\log|e^t| - \frac{1}{2}\log|e^{2t}+e^t+1|-\frac{1}{\sqrt3}\arctan \left(\frac{2e^t+1}{\sqrt3}\right)+C \\&=t - \frac{1}{2}\log|e^{2t}+e^t+1|-\frac{1}{\sqrt3}\arctan \left(\frac{2e^t+1}{\sqrt3}\right)+C\end{align*}
Q26Stage 3

Evaluate ∫1+ex dx\displaystyle\int\sqrt{1+e^x}~\dee{x} using partial fraction.

Hint

Try the substitution u=1+exu=\sqrt{1+e^x}. You'll need to do long division before you can use partial fraction decomposition.

Answer

21+ex+log⁡∣1+ex−11+ex+1∣+C\displaystyle2\sqrt{1+e^x}+\log\left| \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|+C

Full solution

  • We use the substitution u=1+exu=\sqrt{1+e^x}.
    Then du=ex21+exdx\dee{u} = \dfrac{e^x}{2\sqrt{1+e^x}}\dee{x}, so dx=2uu2−1du\dee{x}=\dfrac{2u}{u^2-1}\dee{u}.

    ∫1+ex dx=∫u⋅2uu2−1du=∫2u2u2−1du=∫2(u2−1)+2u2−1du=∫(2+2u2−1)du\begin{align*}\int\sqrt{1+e^x}~\dee{x}&=\int u\cdot\dfrac{2u}{u^2-1}\dee{u} = \int \dfrac{2u^2}{u^2-1}\dee{u} \\ &= \int \dfrac{2(u^2-1)+2}{u^2-1}\dee{u} = \int\left(2+ \dfrac{2}{u^2-1}\right)\dee{u}\end{align*}

    We use a partial fraction decomposition on the fractional part of the integrand.

    2u2−1=2(u−1)(u+1)=Au−1+Bu+1=(A+B)u+(A−B)(u−1)(u+1)A+B=0,A−B=2A=1,B=−1∫1+ex dx=∫(2+2u2−1)du=∫(2+1u−1−1u+1) du=2u+log⁡∣u−1∣−log⁡∣u+1∣+C=2u+log⁡∣u−1u+1∣+C=21+ex+log⁡∣1+ex−11+ex+1∣+C\begin{align*}&\frac{2}{u^2-1} = \frac{2}{(u-1)(u+1)}=\frac{A}{u-1}+\frac{B}{u+1}=\frac{(A+B)u+(A-B)}{(u-1)(u+1)}\\ & A+B=0,\quad A-B=2\\ &\color{red} A=1,\quad B=-1\\ \int\sqrt{1+e^x}~\dee{x}&=\int\left(2+ \dfrac{2}{u^2-1}\right)\dee{u} =\int\left( 2 + \frac{1}{u-1} - \frac{1}{u+1}\right)~\dee{u}\\ &=2u+\log|u-1|-\log|u+1|+C=2u+\log\left| \frac{u-1}{u+1}\right|+C\\ &=2\sqrt{1+e^x}+\log\left| \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|+C\end{align*}

  • It might not occur to us right away to use the fruitful substitution in Solution 1. More realistically, we might start with the “inside function," u=1+exu=1+e^x. Then du=ex dx\dee{u}=e^x\,\dee{x}, so dx=1u−1du\dee{x} = \frac{1}{u-1}\dee{u}.

    ∫1+ex dx=∫uu−1du\begin{align*}\int\sqrt{1+e^x}~\dee{x}&=\int\frac{ \sqrt{u}}{u-1}\dee{u}\end{align*}

    This isn't quite a rational function, because we have a square root on top. If we could turn it into a rational function, we could use partial fraction. To that end, let w=uw=\sqrt{u}, dw=12udu\dee{w} = \frac{1}{2\sqrt{u}}\dee{u}, so du=2wdw\dee{u}=2w\dee{w}.

    =∫ww2−12wdw=∫2w2w2−1dw=∫2(w2−1)+2w2−1dw=∫2+2w2−1dw\begin{align*}&=\int \frac{w}{w^2-1}2w\dee{w}=\int \frac{2w^2}{w^2-1}\dee{w}\\ &=\int \frac{2(w^2-1)+2}{w^2-1}\dee{w} = \int 2 + \frac{2}{w^2-1}\dee{w}\end{align*}

    Now we can use partial fraction decomposition.

    2w2−1=2(w−1)(w+1)=Aw−1+Bw+1=(A+B)w+(A−B)(w−1)(w+1)A+B=0,A−B=2A=1,B=−1\begin{align*}& \frac{2}{w^2-1} = \frac{2}{(w-1)(w+1)} = \frac{A}{w-1}+\frac{B}{w+1} = \frac{(A+B)w+(A-B)}{(w-1)(w+1)}\\ & A+B=0,\quad A-B=2\\ &\color{red} A=1,\quad B=-1\end{align*}

    This allows us to antidifferentiate.

    ∫1+ex dx=∫(2+2w2−1)dw=∫(2+1w−1−1w+1) dw=2w+log⁡∣w−1∣−log⁡∣w+1∣+C=2w+log⁡∣w−1w+1∣+C=2u+log⁡∣u−1u+1∣+C=21+ex+log⁡∣1+ex−11+ex+1∣+C\begin{align*}\int\sqrt{1+e^x}~\dee{x}&= \int \left(2 + \frac{2}{w^2-1}\right)\dee{w}=\int \left(2 + \frac{1}{w-1}-\frac{1}{w+1}\right)~\dee{w}\\ &= 2w + \log|w-1| - \log|w+1|+C\\ &=2w+\log\left| \frac{w-1}{w+1}\right|+C\\ &=2\sqrt{u}+\log\left| \frac{\sqrt{u}-1}{\sqrt{u}+1}\right|+C\\ &=2\sqrt{1+e^x}+\log\left| \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|+C\end{align*}

Remark: we also evaluated this integral using trigonometric substitution in Section 1.9, Question 26. In that question, we found the antiderivative to be 21+ex+2log⁡∣1−1+ex∣−x+C2\sqrt{1+e^x}+2\log\left| 1-\sqrt{1+e^x} \right|-x+C. These expressions are equivalent:

log⁡∣1+ex−11+ex+1∣=log⁡∣1+ex−1∣+log⁡∣11+ex+1∣=log⁡∣1+ex−1∣+log⁡∣(11+ex+1)(1−1+ex1−1+ex)∣=log⁡∣1+ex−1∣+log⁡∣1−1+ex1−(1+ex)∣=log⁡∣1+ex−1∣+log⁡∣1−1+ex−ex∣=log⁡∣1+ex−1∣+log⁡∣1−1+ex∣−log⁡∣−ex∣=2log⁡∣1+ex−1∣−x\begin{align*} \log\left| \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|&= \log\left| \sqrt{1+e^x}-1\right|+\log\left| \frac{1}{\sqrt{1+e^x}+1}\right|\\ &=\log\left| \sqrt{1+e^x}-1\right|+\log\left| \left(\frac{1}{\sqrt{1+e^x}+1}\right)\left(\frac{1-\sqrt{1+e^x}}{1-\sqrt{1+e^x}}\right)\right|\\ &=\log\left| \sqrt{1+e^x}-1\right|+\log\left|\frac{1-\sqrt{1+e^x}}{1-(1+e^x)}\right|\\ &=\log\left| \sqrt{1+e^x}-1\right|+\log\left|\frac{1-\sqrt{1+e^x}}{-e^x}\right|\\ &=\log\left| \sqrt{1+e^x}-1\right|+\log\left|1-\sqrt{1+e^x}\right|-\log|-e^x|\\ &=2\log\left| \sqrt{1+e^x}-1\right|-x \end{align*}
Q27Stage 3Past exam · 1997D

The region RR is the portion of the first quadrant where 3≤x≤43\le x\le 4 and 0≤y≤1025−x20\le y\le\dfrac{10}{\sqrt{25-x^2}}.

  1. Sketch the region RR.

  2. Determine the volume of the solid obtained by revolving RR around the xx–axis.

  3. Determine the volume of the solid obtained by revolving RR around the yy–axis.

Hint

The mechanically easiest way to answer part (c) uses the method of cylindrical shells, which we have not covered. The method of washers also works, but requires you have enough patience and also to have a good idea what RR looks like. So look at the sketch in part (a) very carefully when identifying the left endpoints of your horizontal strips.

Answer

(a)

The region RR is

Figure from prob_s1.10, line 1455

Figure from prob_s1.10, line 1455

(b) 10πlog⁡94=20πlog⁡32\displaystyle10\pi\log\frac{9}{4}=20\pi\log\frac{3}{2} (c) 20π20\pi

Full solution

(a) Let's graph y=1025−x2y=\dfrac{10}{\sqrt{25-x^2}}. We start with the endpoints: (3,52)(3,\frac{5}{2}) and (4,103)(4,\frac{10}{3}). Then we consider the first derivative:

ddx{1025−x2}=10x25−x23\diff{}{x}\left\{\frac{10}{\sqrt{25-x^2}}\right\} = \frac{10x}{\sqrt{25-x^2}^3}

Over the interval [3,4][3,4], this is always positive, so our function is increasing over the entire interval. The second derivative,

d2dx2{1025−x2}=ddx{10x25−x23}=10(2x2+25)25−x25,\ddiff{2}{}{x}\left\{\frac{10}{\sqrt{25-x^2}}\right\}=\diff{}{x}\left\{\frac{10x}{\sqrt{25-x^2}^3}\right\} = \frac{10(2x^2+25)}{\sqrt{25-x^2}^5},

is always positive, so our function is concave up over the entire interval. So, the region RR is:

Figure from prob_s1.10, line 1455

Figure from prob_s1.10, line 1455

(b) Let V1\cV_1 be the solid obtained by revolving RR about the xx–axis. The portion of V1\cV_1 with xx–coordinate between xx and x+dxx+\dee{x} is obtained by rotating the red vertical strip in the figure on the left below about the xx–axis. That portion is a disk of radius 1025−x2\frac{10}{\sqrt{25-x^2}} and thickness dx\dee{x}. The volume of this disk is π(1025−x2)2 dx\pi\big(\frac{10}{\sqrt{25-x^2}}\big)^2\,\dee{x}. So the total volume of V1\cV_1 is

∫34π(1025−x2)2 dx=100π∫34125−x2 dx=100π∫341(5−x)(5+x) dx=10π∫34(15−x+15+x) dx=10π[−log⁡(5−x)+log⁡(5+x)]34=10π[−log⁡1+log⁡9+log⁡2−log⁡8]=10πlog⁡94=20πlog⁡32\begin{align*} \int_3^4\pi{\Big(\frac{10}{\sqrt{25-x^2}}\Big)}^2\ \dee{x} &=100\pi\int_3^4\frac{1}{25-x^2}\ \dee{x} =100\pi\int_3^4\frac{1}{(5-x)(5+x)}\ \dee{x} \\ &=10\pi\int_3^4\Big(\frac{1}{5-x}+\frac{1}{5+x}\Big)\ \dee{x} =10\pi\Big[-\log(5-x)+\log(5+x)\Big]_3^4 \\ &=10\pi\Big[-\log1+\log9+\log2-\log8\Big] =10\pi\log\frac{9}{4}=20\pi\log\frac{3}{2} \end{align*}

Figure from prob_s1.10, line 1470

Figure from prob_s1.10, line 1470

Figure from prob_s1.10, line 1470

Figure from prob_s1.10, line 1470

(c) We'll use horizontal washers as in Example 1.6.5 of the CLP-2 text.

  • We cut R\cR into thin horizontal strips of width dy\dee{y} as in the figure on the right above.

  • When we rotate R\cR about the yy–axis, each strip sweeps out a thin washer

    • whose outer radius is rout=4r_{out}=4, and

    • whose inner radius is rin=25−100y2r_{in}= \sqrt{25-\frac{100}{y^2}} when y≥1025−32=104=52y\ge \frac{10}{\sqrt{25-3^2}} = \frac{10}{4} =\frac{5}{2} (see the red strip in the figure on the right above), and whose inner radius is rin=3r_{in}= 3 when y≤52y\le \frac{5}{2} (see the blue strip in the figure on the right above) and

    • whose thickness is dy\dee{y} and hence

    • whose volume is π(rout2−rin2)dy=π(100y2−9)dy\pi(r_{out}^2 - r_{in}^2)\dee{y} = \pi\big(\frac{100}{y^2}-9\big)\dee{y} when y≥52y\ge \frac{5}{2} and whose volume is π(rout2−rin2)dy=7π dy\pi(r_{out}^2 - r_{in}^2)\dee{y} = 7 \pi\,\dee{y} when y≤52y\le \frac{5}{2} and

  • As our bottommost strip is at y=0y=0 and our topmost strip is at y=103y=\frac{10}{3} (since at the top x=4x=4 and y=1025−x2=1025−42=103y= \frac{10}{\sqrt{25-x^2}} =\frac{10}{\sqrt{25-4^2}} =\frac{10}{3}), the volume is

    ∫5/210/3π(100y2−9)dy+∫05/27π dy=π[−100y−9y]5/210/3+352π=π[−30+40−30+452]+352π=20π\begin{align*} &\int _{5/2}^{10/3} \pi\Big(\frac{100}{y^2}-9\Big)\dee{y} +\int _ 0^{5/2}7 \pi\,\dee{y} \\ &=\pi{\Big[-\frac{100}{y}-9y\Big]}_{5/2}^{10/3} +\frac{35}{2}\pi\\ &=\pi \Big[-30+40-30+\frac{45}{2}\Big] +\frac{35}{2}\pi\\ &=20\pi \end{align*}
Q28Stage 3

Find the area of the finite region bounded by the curves y=43+x2y=\dfrac{4}{3+x^2}, y=2x(x+1)y=\dfrac{2}{x(x+1)}, x=14x=\dfrac14, and x=3x=3.

Hint

You'll need to use two regions, because the curves cross.

Answer

2log⁡53+43arctan⁡143\displaystyle2\log\frac53+\frac{4}{\sqrt3}\arctan\frac{1}{4\sqrt3}

Full solution

In order to find the area between the curves, we need to know which one is on top, and which on the bottom. Let's start by finding where they meet.

43+x2=2x(x+1)2x2+2x=3+x2x2+2x−3=0(x−1)(x+3)=0\begin{align*} \dfrac{4}{3+x^2}&=\frac{2}{x(x+1)}\\ 2x^2+2x&=3+x^2\\ x^2+2x-3&=0\\ (x-1)(x+3)&=0 \end{align*}

In the interval [14,3][\frac14,3], the curves only meet at x=1x=1. So, to find which is on top and on bottom in the intervals [14,1)[\frac14,1) and (1,3](1,3], it suffices to check some point in each interval.

xx43+x2\frac{4}{3+x^2}2x(x+1)\frac{2}{x(x+1)}Top:
1/21/216/1316/138/38/32x(x+1)\frac{2}{x(x+1)}
224/74/71/31/343+x2\frac{4}{3+x^2}

So, 2x(x+1)\frac{2}{x(x+1)} is the top function when 14≤x<1\frac{1}{4}\leq x < 1, and 43+x2\frac{4}{3+x^2} is the top function when 1<x≤31<x \leq 3. Then the area we want to find is:

Area=∫141(2x(x+1)−43+x2) dx+∫13(43+x2−2x(x+1)) dx\text{Area}=\int_{\frac14}^1 \left( \frac{2}{x(x+1)} - \frac{4}{3+x^2}\right)~\dee{x}+ \int_{1}^3 \left( \frac{4}{3+x^2} -\frac{2}{x(x+1)} \right)~\dee{x}

We'll need to antidifferentiate both these functions. We can antidifferentiate 2x(x+1)\dfrac{2}{x(x+1)} using partial fraction decomposition.

2x(x+1)=Ax+Bx+1=(A+B)x+Ax(x+1)A=2,B=−2∫2x(x+1) dx=∫(2x−2x+1) dx=2log⁡∣x∣−2log⁡∣x+1∣+C=2log⁡∣xx+1∣+C\begin{align*} \frac{2}{x(x+1)}&=\frac{A}{x}+\frac{B}{x+1} = \frac{(A+B)x+A}{x(x+1)}\\ \color{red}A&\color{red}=2,\quad B=-2\\ \int\frac{2}{x(x+1)}~\dee{x}&=\int\left(\frac{2}{x}-\frac{2}{x+1}\right)~\dee{x}=2\log|x|-2\log|x+1|+C\\ &=2\log\left| \frac{x}{x+1}\right|+C \end{align*}

We can antidifferentiate 43+x2\dfrac{4}{3+x^2} using the substitution u=x3u=\frac{x}{\sqrt3}, du=13 dx\dee{u}=\frac{1}{\sqrt3}~\dee{x}.

∫43+x2 dx=∫43(1+(x3)2) dx=∫433(1+u2) du=43arctan⁡u+C=43arctan⁡(x3)+C\begin{align*} \int\frac{4}{3+x^2}~\dee{x}&= \int\frac{4}{3\left(1+\left(\frac{x}{\sqrt3}\right)^2\right)}~\dee{x}= \int\frac{4\sqrt3}{3\left(1+u^2\right)}~\dee{u}\\ &=\frac{4}{\sqrt3}\arctan u +C=\frac{4}{\sqrt3}\arctan \left(\frac{x}{\sqrt3}\right) +C \end{align*}

Now, we can find our area.

Area=∫141(2x(x+1)−43+x2) dx+∫13(43+x2−2x(x+1)) dx=[2log⁡∣xx+1∣−43arctan⁡(x3)]1/41+[43arctan⁡(x3)−2log⁡∣xx+1∣]13=(2log⁡12−43⋅π6−2log⁡15+43arctan⁡143)+=(43⋅π3−2log⁡34−43⋅π6+2log⁡12)=2log⁡53+43arctan⁡143\begin{align*} \text{Area}&=\int_{\frac14}^1 \left( \frac{2}{x(x+1)} - \frac{4}{3+x^2}\right)~\dee{x}+ \int_{1}^3 \left( \frac{4}{3+x^2} -\frac{2}{x(x+1)} \right)~\dee{x}\\ &=\left[2\log\left|\frac{x}{x+1}\right| - \frac{4}{\sqrt3}\arctan\left(\frac{x}{\sqrt3}\right)\right] _{1/4}^1+ \left[ \frac{4}{\sqrt3}\arctan\left(\frac{x}{\sqrt3}\right)-2\log\left|\frac{x}{x+1}\right|\right]_1^3\\ &=\left(2\log\frac{1}{2}- \frac{4}{\sqrt3}\cdot\frac{\pi}{6} - 2\log\frac{1}{5}+\frac{4}{\sqrt3}\arctan\frac{1}{4\sqrt3}\right)+\\ &\hphantom{=} \left(\frac{4}{\sqrt3}\cdot\frac{\pi}{3} - 2\log\frac34 - \frac{4}{\sqrt3}\cdot \frac{\pi}{6}+2\log\frac{1}{2}\right)\\ &=2\log\frac53+\frac{4}{\sqrt3}\arctan\frac{1}{4\sqrt3} \end{align*}
Q29Stage 3

Let F(x)=∫1x1t2−9dtF(x) = \displaystyle\int_1^x \frac{1}{t^2-9} \dee{t}.

  1. Give a formula for F(x)F(x) that does not involve an integral.

  2. Find F′(x)F'(x).

Hint

For (b), use the Fundamental Theorem of Calculus Part 1.

Answer

(a) 16(log⁡∣2⋅x−3x+3∣)\displaystyle\frac{1}{6}\left(\log\left| 2\cdot\frac{x-3}{x+3} \right|\right) (b) F′(x)=1x2−9F'(x) = \frac{1}{x^2-9}

Full solution

(a) To antidifferentiate 1t2−9\dfrac{1}{t^2-9}, we use a partial fraction decomposition.

1t2−9=1(t−3)(t+3)=At−3+Bt+3=(A+B)t+3(A−B)(t−3)(t+3)A+B=0,A−B=13A=16,B=−16F(x)=∫1x1t2−9 dx=∫1x(1/6t−3−1/6t+3) dx=[16log⁡∣t−3∣−16log⁡∣t+3∣]1x=(16log⁡∣x−3∣−16log⁡∣x+3∣−16log⁡2+16log⁡4)=16(log⁡∣2⋅x−3x+3∣)\begin{align*} \frac{1}{t^2-9}&=\frac{1}{(t-3)(t+3)} = \frac{A}{t-3}+\frac{B}{t+3}=\frac{(A+B)t+3(A-B)}{(t-3)(t+3)}\\ A+B&=0,\quad A-B = \frac{1}{3}\\ \color{red}A&\color{red}=\frac{1}{6},\quad B = -\frac{1}{6}\\ F(x)&=\int_1^x\frac{1}{t^2-9}~\dee{x} = \int_1^x \left(\frac{1/6}{t-3}-\frac{1/6}{t+3}\right)~\dee{x}\\ &=\left[\frac{1}{6}\log|t-3| - \frac{1}{6}\log|t+3|\right]_1^x\\ &=\left(\frac{1}{6}\log|x-3| - \frac{1}{6}\log|x+3| -\frac{1}{6}\log2 +\frac{1}{6}\log4\right)\\ &=\frac{1}{6}\left(\log\left| 2\cdot\frac{x-3}{x+3} \right|\right) \end{align*}

(b) Rather than differentiate our answer from (a), we use the Fundamental Theorem of Calculus Part 1 to conclude

F′(x)=ddx{∫1x1t2−9dt}=1x2−9F'(x) = \diff{}{x}\left\{ \displaystyle\int_1^x \frac{1}{t^2-9} \dee{t}\right\} = \frac{1}{x^2-9}

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From the UBC Math 101 open textbook project, © Joel Feldman, Andrew Rechnitzer, Elyse Yeager and others. Licensed CC BY-NC-SA 4.0; this HTML adaptation is released under the same licence. Full attribution.