1.10 Partial Fractions

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

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

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

$\displaystyle\frac{A}{x-1}+\frac{B}{(x-1)^2}+\frac{C}{x+1}+\frac{D}{(x+1)^2}+\frac{Ex+F}{x^2+1}$

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

$3$

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

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

(b) $\displaystyle\dfrac{15x^4+6x^3+34x^2+4x+20}{5x^2+2x+8} = 3x^2+2+\frac{4}{5x^2+2x+8}$

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

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

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

(b) $x^4-3x^2-5=\displaystyle\left(x+\sqrt{\frac{3+\sqrt{29}}{2}}\right)\left(x-\sqrt{\frac{3+\sqrt{29}}{2}}\right)\left(x^2+\frac{\sqrt{29}-3}{2}\right)$

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

Why do we do partial fraction decomposition at all?

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 $u=(ax+b)$. It's less clear how to find the antiderivative of (2).

What is the title of this section?

$\displaystyle\log\frac{4}{3}$

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

$-\dfrac{1}{x}-\arctan x+C$

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

$4 \log |x-3| - 2 \log (x^2 + 1) + C$

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

$F(x) = \log |x-2| + \log (x^2+4) + 2\arctan (x/2) + D$

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

$-2\log|x-3|+3\log|x+2|+C$

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

$-9\log|x+2|+14\log|x+3| +C$

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.

$\displaystyle5x+\frac{1}{2}\log|x-1| - \frac{7}{2}\log|x+1|+C$

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

$\displaystyle x-\frac{2}{x}+\frac{5}{2}\arctan (2x) +C$

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

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

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

$\displaystyle-\frac{1}{2}\log|x-2| + \frac{1}{2}\log|x+2| + \frac{3}{2}\log|2x-1|+C$

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

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

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

$\displaystyle\frac{1}{2}\log\left| \frac{1-\cos x}{1+\cos x}\right|+C$

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

$\displaystyle\frac{-\cos x}{2\sin^2 x} + \frac{1}{4}\log\left|\frac{1-\cos x}{1+\cos x}\right|+C$

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

$\displaystyle3\log 2 + \frac{1}{2}+\frac{2}{\sqrt{15}}\left(\arctan\left(\frac{7}{\sqrt{15}}\right)-\arctan\left(\frac{9}{\sqrt{15}}\right)\right)$

Review Question 20 in Section 1.9 for antidifferentiation tips.

$\displaystyle=\frac{9}{4\sqrt2}\arctan \left(\frac{x}{\sqrt2}\right)-\frac{2+3x}{4(x^2+2)} +C$

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

$\displaystyle\frac{3}{8}\arctan x + \frac{3x^3+5x}{8(1+x^2)^2}+C$

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

$\displaystyle\frac{3}{2}x^2+\frac{1}{\sqrt{5}}\arctan \left(\frac{x}{\sqrt5}\right)+ \frac{3}{2}\log|x^2+5|-\frac{3}{2x^2+10}+C$

$\cos^2\theta = 1-\sin^2\theta$

$\displaystyle\log\left| \frac{\sin\theta-1}{\sin\theta-2}\right|+C$

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

$\displaystyle t - \frac{1}{2}\log|e^{2t}+e^t+1|-\frac{1}{\sqrt3}\arctan \left(\frac{2e^t+1}{\sqrt3}\right)+C$

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

$\displaystyle2\sqrt{1+e^x}+\log\left| \frac{\sqrt{1+e^x}-1}{\sqrt{1+e^x}+1}\right|+C$

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 $R$ looks like. So look at the sketch in part (a) very carefully when identifying the left endpoints of your horizontal strips.

(a)

The region $R$ is

(b) $\displaystyle10\pi\log\frac{9}{4}=20\pi\log\frac{3}{2}$ (c) $20\pi$

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

$\displaystyle2\log\frac53+\frac{4}{\sqrt3}\arctan\frac{1}{4\sqrt3}$

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

(a) $\displaystyle\frac{1}{6}\left(\log\left| 2\cdot\frac{x-3}{x+3} \right|\right)$ (b) $F'(x) = \frac{1}{x^2-9}$