Integration using partial fractions

​​In a nutshell

Partial fractions can be used to integrate algebraic fractions.

Example 1

Evaluate $$\int \dfrac{3x-3}{x^2-x-2}\,dx$$​

Split up the algebraic fractions into partial fractions:

​$$\begin{aligned}\dfrac{3x-3}{x^2-x-2} &\equiv \dfrac{3x-3}{(x-2)(x+1)} \\&\equiv \dfrac{A}{x-2} +\dfrac{B}{x+1}\\ 3x-3 &\equiv A(x+1)+B(x-2) \end{aligned}$$​​

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

​$$\dfrac{3x-3}{x^2-x-2} \equiv \dfrac{1}{x-2}+\dfrac{2}{x+1}$$​​

Perform the integration:

​$$\begin{aligned}\int \dfrac{3x-3}{x^2-x-2}\,dx &= \int \left( \dfrac{1}{x-2} + \dfrac{2}{x+1}\right)\, dx\\&=\int \dfrac{1}{x-2}\,dx + \int \dfrac{2}{x+1}\,dx\\&=\ln\vert x-2\vert + 2\ln\vert x+1\vert +C \end{aligned}$$​​

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​$$\int \dfrac{3x-3}{x^2-x-2}\,dx =\underline{\ln\vert x-2\vert + 2\ln\vert x+1\vert +C }$$​​

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 $$\int k\dfrac{f'(x)}{f(x)}\,dx$$, 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 $$\int \dfrac{x^3+2x-1}{x-3}\,dx$$.

Use polynomial long division to simplify the integral:

$$\begin{aligned} \dfrac{x^3+2x-1}{x-3}&\equiv \dfrac{x^3-3x^2+3x^2-9x+11x-33+32}{x-3}\\&\equiv\dfrac{x^2(x-3)+3x(x-3)+11(x-3)+32}{x-3}\\&\equiv x^2+3x+11+\dfrac{32}{x-3} \end{aligned}$$​​

Perform the integration:

$$\begin{aligned}\int \dfrac{x^3+2x-1}{x-3}\,dx &\equiv \int \left(x^2+3x+11+\dfrac{32}{x-3}\right)\,dx\\&=\dfrac13x^3+\dfrac32x^2+11x+32\ln\vert x-3\vert+C \end{aligned}$$​​

$$\int \dfrac{x^3+2x-1}{x-3}\,dx =\underline{\dfrac13x^3+\dfrac32x^2+11x+32\ln\vert x-3\vert+C}$$​​

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