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Integration II

Integration using partial fractions

Integration using partial fractions

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Explainer Video

Tutor: Bilal

Summary

Integration using partial fractions

​​In a nutshell

Partial fractions can be used to integrate algebraic fractions.


Example 1

Evaluate ∫3x−3x2−x−2 dx\int \dfrac{3x-3}{x^2-x-2}\,dx∫x2−x−23x−3​dx​


Split up the algebraic fractions into partial fractions:

​3x−3x2−x−2≡3x−3(x−2)(x+1)≡Ax−2+Bx+13x−3≡A(x+1)+B(x−2)\begin{aligned}\dfrac{3x-3}{x^2-x-2} &\equiv \dfrac{3x-3}{(x-2)(x+1)} \\&\equiv \dfrac{A}{x-2} +\dfrac{B}{x+1}\\ 3x-3 &\equiv A(x+1)+B(x-2) \end{aligned}x2−x−23x−3​3x−3​≡(x−2)(x+1)3x−3​≡x−2A​+x+1B​≡A(x+1)+B(x−2)​​​


​x=2→3(2)−3=3A→A=1x=−1→3(−1)−3=−3B→B=2\begin{aligned}x&=2 \rightarrow 3(2)-3 = 3A\rightarrow A=1\\x&=-1 \rightarrow 3(-1)-3 = -3B \rightarrow B=2 \end{aligned}xx​=2→3(2)−3=3A→A=1=−1→3(−1)−3=−3B→B=2​​​


​3x−3x2−x−2≡1x−2+2x+1\dfrac{3x-3}{x^2-x-2} \equiv \dfrac{1}{x-2}+\dfrac{2}{x+1}x2−x−23x−3​≡x−21​+x+12​​​


Perform the integration:

​∫3x−3x2−x−2 dx=∫(1x−2+2x+1) dx=∫1x−2 dx+∫2x+1 dx=ln⁡∣x−2∣+2ln⁡∣x+1∣+C\begin{aligned}\int \dfrac{3x-3}{x^2-x-2}\,dx &= \int \left( \dfrac{1}{x-2} + \dfrac{2}{x+1}\right)\, dx\\&=\int \dfrac{1}{x-2}\,dx + \int \dfrac{2}{x+1}\,dx\\&=\ln\vert x-2\vert + 2\ln\vert x+1\vert +C \end{aligned}∫x2−x−23x−3​dx​=∫(x−21​+x+12​)dx=∫x−21​dx+∫x+12​dx=ln∣x−2∣+2ln∣x+1∣+C​​​

​​

​∫3x−3x2−x−2 dx=ln⁡∣x−2∣+2ln⁡∣x+1∣+C‾\int \dfrac{3x-3}{x^2-x-2}\,dx =\underline{\ln\vert x-2\vert + 2\ln\vert x+1\vert +C }∫x2−x−23x−3​dx=ln∣x−2∣+2ln∣x+1∣+C​​​


Note: Using partial fractions should be a last resort. Always check if you can use integration by substitution, or check if the integral is of the form ∫kf′(x)f(x) dx\int k\dfrac{f'(x)}{f(x)}\,dx∫kf(x)f′(x)​dx, which would allow for integration via the reverse chain rule.


Algebraic division

It may also be necessary to use polynomial long division to simplify an integral.


Example 2

Evaluate ∫x3+2x−1x−3 dx\int \dfrac{x^3+2x-1}{x-3}\,dx∫x−3x3+2x−1​dx.


Use polynomial long division to simplify the integral:

x3+2x−1x−3≡x3−3x2+3x2−9x+11x−33+32x−3≡x2(x−3)+3x(x−3)+11(x−3)+32x−3≡x2+3x+11+32x−3\begin{aligned} \dfrac{x^3+2x-1}{x-3}&\equiv \dfrac{x^3-3x^2+3x^2-9x+11x-33+32}{x-3}\\&\equiv\dfrac{x^2(x-3)+3x(x-3)+11(x-3)+32}{x-3}\\&\equiv x^2+3x+11+\dfrac{32}{x-3} \end{aligned}x−3x3+2x−1​​≡x−3x3−3x2+3x2−9x+11x−33+32​≡x−3x2(x−3)+3x(x−3)+11(x−3)+32​≡x2+3x+11+x−332​​​​


Perform the integration:

∫x3+2x−1x−3 dx≡∫(x2+3x+11+32x−3) dx=13x3+32x2+11x+32ln⁡∣x−3∣+C\begin{aligned}\int \dfrac{x^3+2x-1}{x-3}\,dx &\equiv \int \left(x^2+3x+11+\dfrac{32}{x-3}\right)\,dx\\&=\dfrac13x^3+\dfrac32x^2+11x+32\ln\vert x-3\vert+C \end{aligned}∫x−3x3+2x−1​dx​≡∫(x2+3x+11+x−332​)dx=31​x3+23​x2+11x+32ln∣x−3∣+C​​​


∫x3+2x−1x−3 dx=13x3+32x2+11x+32ln⁡∣x−3∣+C‾\int \dfrac{x^3+2x-1}{x-3}\,dx =\underline{\dfrac13x^3+\dfrac32x^2+11x+32\ln\vert x-3\vert+C} ∫x−3x3+2x−1​dx=31​x3+23​x2+11x+32ln∣x−3∣+C​​​


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Operations with algebraic fractions

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Operations with algebraic fractions

Partial fractions

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Partial fractions

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FAQs - Frequently Asked Questions

How do you integrate algebraic fractions where the numerator has a higher power than the denominator?

Use polynomial long division.

When do you use partial fractions to integrate?

Only use partial fractions once you're sure that reverse chain rule or integration by substitution doesn't work.

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