# How to Use Partial Fraction Decomposition for Integration

Partial fraction decomposition has a long name, but it is an easy and simple method. You use the factors in the denominator to create new and nicer-looking fractions. When the new fractions are set, the integration becomes much easier. Nice!

Rule

### InstructionsforPartialFractionDecomposition

1.
Factorize the denominator by finding the zeros.
2.
Intermediate calculation (see the box below):
a)
Set the expression equal to a sum of fractions and choose the constants $A,B,C,\dots$ in the numerator—one letter for each fraction.
b)
Multiply by the common denominator.
c)
Sort the different terms individually.
d)
Create a system of equations and solve for $A,B,C,\dots$
3.
Insert the result into the integral and integrate. Yay!

Example 1

Compute $\int \frac{3x}{{x}^{2}-x-2}\phantom{\rule{0.17em}{0ex}}dx$

$\begin{array}{llll}\hfill \int \frac{3x}{{x}^{2}-x-2}\phantom{\rule{0.17em}{0ex}}dx& =\int \frac{3x}{\left(x-2\right)\left(x+1\right)}\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \stackrel{\ast }{=}\int \frac{2}{x-2}+\frac{1}{x+1}\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|x-2|+\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|x+1|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

$\begin{array}{llll}\hfill \int \frac{3x}{{x}^{2}-x-2}\phantom{\rule{0.17em}{0ex}}dx& =\int \frac{3x}{\left(x-2\right)\left(x+1\right)}\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & \stackrel{\ast }{=}\int \frac{2}{x-2}+\frac{1}{x+1}\phantom{\rule{0.17em}{0ex}}dx\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\\ \hfill & =2\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|x-2|+\mathrm{ln}\phantom{\rule{-0.17em}{0ex}}|x+1|+C\phantom{\rule{2em}{0ex}}& \hfill & \phantom{\rule{2em}{0ex}}\end{array}$

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3x (x 2)(x + 1) = A x 2 + B x + 1 3x = A(x + 1) + B(x 2) 3x = Ax + A + Bx 2B 3x + 0 = (A + B)x + (A 2B) Equate terms of the same degree and solve. You then get A + B = 3 and A 2B = 0, which gives A = 2 and B = 1.
3x (x 2)(x + 1) = A x 2 + B x + 1 3x = A(x + 1) + B(x 2) 3x = Ax + A + Bx 2B 3x + 0 = (A + B)x + (A 2B) Equate terms of the same degree and solve. You then get A + B = 3 and A 2B = 0, which gives A = 2 and B = 1.