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Topic 7

Partial Fractions

Partial fractions splits a into simple pieces you already know how to integrate. A piece like Ax−a\dfrac{A}{x - a} gives a log⁡\log, and Bx+Dx2+k2\dfrac{Bx + D}{x^2 + k^2} gives a log⁡\log plus an arctan⁡\arctan. It only works when the top's degree is less than the bottom's, so divide first if it is not.

5 kinds of question. Open the one you are stuck on.

1

Integrate a rational function using partial fractions

e.g. “Evaluate ∫P(x)Q(x) dx\int\dfrac{P(x)}{Q(x)}\,dx where QQ factors into different linear factors”
  1. 1
    Factor the denominator completely

    First check that the top's degree is less than the bottom's. If not, divide. Factor a cubic by testing small roots like ±1\pm1, ±2\pm2.

  2. 2
    Write one term Ax−a\dfrac{A}{x - a} per factor
  3. 3
    Find each constant by plugging in a root

    Multiply through by the denominator, then set x=ax = a: every other term becomes 00.

  4. 4
    Integrate each term: Ax−a\dfrac{A}{x - a} gives Alog⁡∣x−a∣A\log|x - a|

    For A2x−1\dfrac{A}{2x - 1} you get A2log⁡∣2x−1∣\tfrac{A}{2}\log|2x - 1|.

Remember: Factor, split, plug in roots, log.

See it in a problem3 worked examples

Evaluate ∫5x−4x2−x−2 dx\displaystyle\int\frac{5x - 4}{x^2 - x - 2}\,dx.

  1. 1
    Factor the denominator completely

    x2−x−2=(x−2)(x+1)x^2 - x - 2 = (x - 2)(x + 1); the top has degree 1<21 < 2.

  2. 2
    Write one term Ax−a\dfrac{A}{x - a} per factor

    5x−4(x−2)(x+1)=Ax−2+Bx+1\dfrac{5x - 4}{(x - 2)(x + 1)} = \dfrac{A}{x - 2} + \dfrac{B}{x + 1}.

  3. 3
    Find each constant by plugging in a root

    5x−4=A(x+1)+B(x−2)5x - 4 = A(x + 1) + B(x - 2). x=2x = 2: 6=3A6 = 3A, A=2A = 2. x=−1x = -1: −9=−3B-9 = -3B, B=3B = 3.

  4. 4
    Integrate each term: Ax−a\dfrac{A}{x - a} gives Alog⁡∣x−a∣A\log|x - a|

    2log⁡∣x−2∣+3log⁡∣x+1∣+C2\log|x - 2| + 3\log|x + 1| + C.

Answer:

2log⁡∣x−2∣+3log⁡∣x+1∣+C2\log|x - 2| + 3\log|x + 1| + C

Common mistakes3
  • Dropping the absolute values: log⁡∣x−2∣\log|x - 2|, not log⁡(x−2)\log(x - 2).
  • Skipping the degree check: the method needs the top's degree to be smaller.
  • Forgetting the 1a\tfrac1a when the factor is ax+bax + b.
2

Use long division before partial fractions

e.g. “Rewrite as the sum of a polynomial and a rational function whose numerator has a strictly smaller degree than its denominator”
  1. 1
    Compare degrees: divide if top ≥\ge bottom
  2. 2
    Long-divide: quotient + remainderdenominator+\ \dfrac{\text{remainder}}{\text{denominator}}
  3. 3
    Split the remainder fraction with partial fractions
  4. 4
    Integrate the polynomial and each fraction

Remember: Top too big? Divide first.

See it in a problem2 worked examples

Evaluate ∫x3+2xx2−1 dx\displaystyle\int\frac{x^3 + 2x}{x^2 - 1}\,dx.

  1. 1
    Compare degrees: divide if top ≥\ge bottom

    Degree 3≥23 \ge 2: divide.

  2. 2
    Long-divide: quotient + remainderdenominator+\ \dfrac{\text{remainder}}{\text{denominator}}

    x3+2x=x(x2−1)+3xx^3 + 2x = x(x^2 - 1) + 3x, so x3+2xx2−1=x+3xx2−1\dfrac{x^3 + 2x}{x^2 - 1} = x + \dfrac{3x}{x^2 - 1}.

  3. 3
    Split the remainder fraction with partial fractions

    3x(x−1)(x+1)=Ax−1+Bx+1\dfrac{3x}{(x - 1)(x + 1)} = \dfrac{A}{x - 1} + \dfrac{B}{x + 1}: x=1x = 1 gives A=32A = \tfrac32; x=−1x = -1 gives B=32B = \tfrac32.

  4. 4
    Integrate the polynomial and each fraction

    x22+32log⁡∣x−1∣+32log⁡∣x+1∣+C\dfrac{x^2}{2} + \dfrac32\log|x - 1| + \dfrac32\log|x + 1| + C.

Answer:

x22+32log⁡∣x−1∣+32log⁡∣x+1∣+C\dfrac{x^2}{2} + \dfrac32\log|x - 1| + \dfrac32\log|x + 1| + C

Common mistakes2
  • Decomposing before dividing: the equations for AA and BB then have no solution.
  • Dropping the quotient after dividing: it has to be integrated too.
3

Integrate with partial fractions when a factor is repeated

e.g. “a squared or cubed factor in the denominator: x2x^2, (x−2)2(x - 2)^2, (x+1)3(x + 1)^3”
  1. 1
    Factor and note each repeated factor
  2. 2
    Give (x−a)k(x - a)^k the terms A1x−a+⋯+Ak(x−a)k\dfrac{A_1}{x - a} + \dots + \dfrac{A_k}{(x - a)^k}
  3. 3
    Plug in roots, then compare coefficients for the rest

    Each root gives the constant over its factor's highest power. Matching coefficients, often of x2x^2, gives the rest.

  4. 4
    Integrate: only 1x−a\dfrac{1}{x - a} gives a log

    ∫dx(x−a)k=−1(k−1)(x−a)k−1\displaystyle\int\frac{dx}{(x - a)^k} = -\frac{1}{(k - 1)(x - a)^{k-1}} for k≥2k \ge 2.

Remember: A repeated factor gets every power up to its own.

See it in a problem3 worked examples

Evaluate ∫x+4x(x−2)2 dx\displaystyle\int\frac{x + 4}{x(x - 2)^2}\,dx.

  1. 1
    Factor and note each repeated factor

    The denominator is x(x−2)2x(x - 2)^2; the factor x−2x - 2 appears twice.

  2. 2
    Give (x−a)k(x - a)^k the terms A1x−a+⋯+Ak(x−a)k\dfrac{A_1}{x - a} + \dots + \dfrac{A_k}{(x - a)^k}

    Ax+Bx−2+D(x−2)2\dfrac{A}{x} + \dfrac{B}{x - 2} + \dfrac{D}{(x - 2)^2}.

  3. 3
    Plug in roots, then compare coefficients for the rest

    x+4=A(x−2)2+Bx(x−2)+Dxx + 4 = A(x - 2)^2 + Bx(x - 2) + Dx. x=0x = 0: A=1A = 1. x=2x = 2: 6=2D6 = 2D, D=3D = 3. x2x^2 terms: 0=A+B0 = A + B, B=−1B = -1.

  4. 4
    Integrate: only 1x−a\dfrac{1}{x - a} gives a log

    log⁡∣x∣−log⁡∣x−2∣−3x−2+C\log|x| - \log|x - 2| - \dfrac{3}{x - 2} + C.

Answer:

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

Common mistakes2
  • Writing only B(x−2)2\dfrac{B}{(x - 2)^2} and skipping Ax−2\dfrac{A}{x - 2}: the equations then usually have no solution.
  • Integrating 1(x−2)2\dfrac{1}{(x - 2)^2} as a log: it is −1x−2-\dfrac{1}{x - 2}.
4

Integrate with partial fractions when a quadratic factor is irreducible

e.g. “a quadratic factor that will not factor, like x2+4x^2 + 4 or x2+2x+5x^2 + 2x + 5”
  1. 1
    Check the quadratic: b2−4ac<0b^2 - 4ac < 0 means irreducible
  2. 2
    Give it a linear top: Bx+Dquadratic\dfrac{Bx + D}{\text{quadratic}}
  3. 3
    Find the constants: plug in roots, then match coefficients
  4. 4
    Split the quadratic term into a log⁡\log part and an arctan⁡\arctan part

    ∫x dxx2+k2=12log⁡(x2+k2)\displaystyle\int\frac{x\,dx}{x^2 + k^2} = \tfrac12\log(x^2 + k^2),  ∫dxx2+k2=1karctan⁡xk\ \displaystyle\int\frac{dx}{x^2 + k^2} = \tfrac1k\arctan\tfrac{x}{k}. With an xx term, complete the square first.

Remember: Quadratic on the bottom, linear on the top.

See it in a problem3 worked examples

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

  1. 1
    Check the quadratic: b2−4ac<0b^2 - 4ac < 0 means irreducible

    x2+4x^2 + 4: b2−4ac=−16<0b^2 - 4ac = -16 < 0, so it is .

  2. 2
    Give it a linear top: Bx+Dquadratic\dfrac{Bx + D}{\text{quadratic}}

    Ax−1+Bx+Dx2+4\dfrac{A}{x - 1} + \dfrac{Bx + D}{x^2 + 4}.

  3. 3
    Find the constants: plug in roots, then match coefficients

    2x2+2x+1=A(x2+4)+(Bx+D)(x−1)2x^2 + 2x + 1 = A(x^2 + 4) + (Bx + D)(x - 1). x=1x = 1: 5=5A5 = 5A, A=1A = 1. x2x^2 terms: 2=A+B2 = A + B, B=1B = 1. Constants: 1=4A−D1 = 4A - D, D=3D = 3.

  4. 4
    Split the quadratic term into a log⁡\log part and an arctan⁡\arctan part

    ∫x+3x2+4 dx=12log⁡(x2+4)+32arctan⁡x2\displaystyle\int\frac{x + 3}{x^2 + 4}\,dx = \tfrac12\log(x^2 + 4) + \tfrac32\arctan\tfrac{x}{2}.

Answer:

log⁡∣x−1∣+12log⁡(x2+4)+32arctan⁡x2+C\log|x - 1| + \dfrac12\log(x^2 + 4) + \dfrac32\arctan\dfrac{x}{2} + C

Common mistakes3
  • Giving the quadratic only a constant on top, Bx2+4\dfrac{B}{x^2 + 4}: it needs Bx+DBx + D.
  • Forgetting the 1k\tfrac1k in 1karctan⁡xk\tfrac1k\arctan\tfrac{x}{k}.
  • Factoring x2+4x^2 + 4 as (x+2)(x−2)(x + 2)(x - 2): that is x2−4x^2 - 4.
5

Use a substitution to turn an integral into partial fractions

e.g. “a rational function of exe^{x}, sin⁡θ\sin\theta or tan⁡θ\tan\theta, with the right derivative beside it”
  1. 1
    Substitute to get a rational function of uu

    u=exu = e^{x}: du=ex dxdu = e^{x}\,dx.  u=tan⁡θ\ u = \tan\theta: du=sec⁡2θ dθdu = \sec^2\theta\,d\theta.  u=x\ u = \sqrt{x}: x=u2x = u^2, dx=2u dudx = 2u\,du.

  2. 2
    Factor and decompose in uu

    Divide first if the top's degree is not smaller.

  3. 3
    Integrate each piece
  4. 4
    Substitute back to the original variable

Remember: Substitute until it is a fraction of polynomials.

See it in a problem3 worked examples

Evaluate ∫exe2x−1 dx\displaystyle\int\frac{e^{x}}{e^{2x} - 1}\,dx.

  1. 1
    Substitute to get a rational function of uu

    u=exu = e^{x}, du=ex dxdu = e^{x}\,dx: ∫duu2−1\displaystyle\int\frac{du}{u^2 - 1}.

  2. 2
    Factor and decompose in uu

    1(u−1)(u+1)=1/2u−1−1/2u+1\dfrac{1}{(u - 1)(u + 1)} = \dfrac{1/2}{u - 1} - \dfrac{1/2}{u + 1}.

  3. 3
    Integrate each piece

    12log⁡∣u−1∣−12log⁡∣u+1∣\tfrac12\log|u - 1| - \tfrac12\log|u + 1|.

  4. 4
    Substitute back to the original variable

    12log⁡∣ex−1ex+1∣\tfrac12\log\left|\dfrac{e^{x} - 1}{e^{x} + 1}\right|.

Answer:

12log⁡∣ex−1ex+1∣+C\dfrac12\log\left|\dfrac{e^{x} - 1}{e^{x} + 1}\right| + C

Common mistakes2
  • Forgetting to convert dxdx: with u=xu = \sqrt{x}, dx=2u dudx = 2u\,du adds a factor of 2u2u on top.
  • Leaving uu in the final answer.

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