# How do you integrate 16/((x-1)^2(x+1)^3) using partial fractions?

Aug 11, 2018

3ln(|x+1|)−4/(x+1)−2/(x+1)^2−2/(x−1)−3ln(|x−1|)+C

#### Explanation:

First of all, whoever asked you this question must hate you a lot lol. Okay, so since the denominator is already in factored form, you can set up your partial fraction immediately.

$\frac{16}{{\left(x - 1\right)}^{2} {\left(x + 1\right)}^{3}} = \frac{a}{x - 1} ^ 2 + \frac{b}{x - 1} + \frac{b x + c}{x + 1} ^ 3 + \frac{\mathrm{dx} + e}{x + 1} ^ 2 + \frac{f x + g}{x + 1}$

Multiply both sides by the denominator of the left. This will get rid of all of the denominators on both sides leaving some stuff in the numerators in the terms on the right:

$16 = a {\left(x + 1\right)}^{3} + b \left(x - 1\right) {\left(x + 1\right)}^{3} + \left(c x + d\right) {\left(x - 1\right)}^{2} + \left(\mathrm{dx} + e\right) {\left(x - 1\right)}^{2} \left(x + 1\right) + \left(f x + g\right) {\left(x - 1\right)}^{2} \left(x + 1\right)$

Then set up a system of equations by letting $x$ = various numbers and solving for variables . You will eventually get:

$\frac{16}{{\left(x - 1\right)}^{2} {\left(x + 1\right)}^{3}} = \frac{2}{x - 1} ^ 2 - \frac{3}{x - 1} + \frac{4}{x + 1} ^ 3 + \frac{4}{x + 1} ^ 2 + \frac{3}{x + 1}$

The integral from there should be pretty straight forward.

$\int \frac{2}{x - 1} ^ 2 - \frac{3}{x - 1} + \frac{4}{x + 1} ^ 3 + \frac{4}{x + 1} ^ 2 + \frac{3}{x + 1}$

=

3ln(|x+1|)−4/(x+1)−2/(x+1)^2−2/(x−1)−3ln(|x−1|)+C