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

Apr 8, 2017

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

$= - \frac{1}{2} \ln \left\mid x - 1 \right\mid - \frac{1}{x - 1} + \frac{1}{2} \ln \left\mid x + 1 \right\mid + C$

#### Explanation:

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

$\textcolor{w h i t e}{\frac{2}{{\left(x - 1\right)}^{2} \left(x + 1\right)}} = \frac{A \left(x - 1\right) \left(x + 1\right) + B \left(x + 1\right) + C {\left(x - 1\right)}^{2}}{{\left(x - 1\right)}^{2} \left(x + 1\right)}$

$\textcolor{w h i t e}{\frac{2}{{\left(x - 1\right)}^{2} \left(x + 1\right)}} = \frac{\left(A + C\right) {x}^{2} + \left(B - 2 C\right) x + \left(- A + B + C\right)}{{\left(x - 1\right)}^{2} \left(x + 1\right)}$

Equating coefficients, we find:

$\left\{\begin{matrix}A + C = 0 \\ B - 2 C = 0 \\ - A + B + C = 2\end{matrix}\right.$

Adding all three equations together, we find:

$2 B = 2$

and hence:

$B = 1$

Then from the second equation we find:

$C = \frac{1}{2}$

Then from the first equation we find:

$A = - \frac{1}{2}$

So:

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

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

$= - \frac{1}{2} \ln \left\mid x - 1 \right\mid - \frac{1}{x - 1} + \frac{1}{2} \ln \left\mid x + 1 \right\mid + C$