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

Mar 28, 2017

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

#### Explanation:

We have:

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

Multiply and divide the integrand function by $x$:

int dx/(x(x^2-1)^2) = int (xdx)/(x^2(x^2-1)^2

Substitute now: ${x}^{2} = t$; $2 x \mathrm{dx} = \mathrm{dt}$

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

Add and subtract $t$ to the numerator:

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

using linearity:

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

and simplifying:

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

The first integral can be resolved directly:

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

For the second, we can divide in partial fractions using the same method:

$\int \frac{1}{t \left(t - 1\right)} \mathrm{dt} = \int \frac{1 - t + t}{t \left(t - 1\right)} \mathrm{dt} = \int \frac{1 - t}{t \left(t - 1\right)} \mathrm{dt} + \int \frac{t \mathrm{dt}}{t \left(t - 1\right)}$

$\int \frac{1}{t \left(t - 1\right)} \mathrm{dt} = - \int \frac{\mathrm{dt}}{t} + \int \frac{\mathrm{dt}}{t - 1}$

$\int \frac{1}{t \left(t - 1\right)} \mathrm{dt} = - \ln \left\mid t \right\mid + \ln \left\mid t - 1 \right\mid + C$

Substituting in the expression $\left(1\right)$ above:

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

and undoing the variable substitution:

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

and finally, using the properties of logarithms:

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