# How do you integrate int 1/ [(x-1) ( x+ 1) ^2]  using partial fractions?

May 13, 2016

$\int \frac{1}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \mathrm{dx}$

= $\frac{1}{4} \ln \left(x - 1\right) - \frac{1}{4} \ln \left(x + 1\right) + \frac{1}{2 \left(x + 1\right)} + c$

#### Explanation:

Let us first find partial fractions of $\frac{1}{\left(x - 1\right) {\left(x + 1\right)}^{2}}$ and for this let

$\frac{1}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \Leftrightarrow \frac{A}{x - 1} + \frac{B}{x + 1} + \frac{C}{x + 1} ^ 2$ or

$\frac{1}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \Leftrightarrow \frac{A {\left(x + 1\right)}^{2} + B \left(x - 1\right) \left(x + 1\right) + C \left(x - 1\right)}{\left(x - 1\right) {\left(x + 1\right)}^{2}}$ or

$\frac{1}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \Leftrightarrow \frac{A \left({x}^{2} + 2 x + 1\right) + B \left({x}^{2} - 1\right) + C \left(x - 1\right)}{\left(x - 1\right) {\left(x + 1\right)}^{2}}$ or

$\frac{1}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \Leftrightarrow \frac{\left(A + B\right) {x}^{2} + \left(2 A + C\right) x + \left(A - B - C\right)}{\left(x - 1\right) {\left(x + 1\right)}^{2}}$

Therefore $A + B = 0$ and $2 A + C = 0$ and $A - B - C = 1$

From these $B = - A$, $C = - 2 A$, Hence $A - \left(- A\right) - \left(- 2 A\right) = 1$ or $4 A = 1$

Hence $A = \frac{1}{4}$, $B = - \frac{1}{4}$ and $C = - \frac{1}{2}$

Hence $\int \frac{1}{\left(x - 1\right) {\left(x + 1\right)}^{2}} \mathrm{dx}$

= $\int \left[\frac{1}{4 \left(x - 1\right)} - \frac{1}{4 \left(x + 1\right)} - \frac{1}{2 {\left(x + 1\right)}^{2}}\right] \mathrm{dx}$

= $\frac{1}{4} \ln \left(x - 1\right) - \frac{1}{4} \ln \left(x + 1\right) + \frac{1}{2 \left(x + 1\right)} + c$