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

##### 1 Answer
Dec 25, 2017

The answer is $= - \frac{1}{2} \ln \left(| {x}^{2} - 1 |\right) + \ln \left(| x |\right) - \frac{1}{2 \left({x}^{2} - 1\right)} + C$

#### Explanation:

Reminder

${x}^{2} - 1 = \left(x + 1\right) \left(x - 1\right)$

Let's perform the decomposition into partial fractions

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

$= \frac{A}{x} + \frac{B}{x + 1} ^ 2 + \frac{C}{x + 1} + \frac{D}{x - 1} ^ 2 + \frac{E}{x - 1}$

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

The denominators are the same, compare the numerators

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

Let $x = 0$, $\implies$, $1 = A$

Let $x = 1$, $\implies$, $1 = 4 D$, $\implies$, $D = \frac{1}{4}$

Let $x = - 1$, $\implies$, $1 = - 4 B$, $\implies$, $B = - \frac{1}{4}$

Coefficients of ${x}^{4}$

$0 = A + C + E$, $\implies$, $C + E = - A = - 1$

Coefficients of ${x}^{3}$

$0 = B + D - C + E$, $\implies$, $B + D = C - E = 0$

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

Therefore,

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

So,

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

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