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

Jan 5, 2017

The answer is =-ln(∣x∣)+1/2ln(x^2+1)+arctanx+C

#### Explanation:

Let's factorise the denominator

${x}^{3} + x = x \left({x}^{2} + 1\right)$

So,

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

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

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

Therefore,

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

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

Coefficients of ${x}^{2}$

$0 = A + B$, $\implies$, $B = 1$

Coefficients of $x$

$1 = C$

So,

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

Therefore,

$\int \frac{\left(x - 1\right) \mathrm{dx}}{{x}^{3} + x} = \int \frac{- 1 \mathrm{dx}}{x} + \int \frac{\left(x + 1\right) \mathrm{dx}}{{x}^{2} + 1}$

$= \int \frac{- 1 \mathrm{dx}}{x} + \int \frac{x \mathrm{dx}}{{x}^{2} + 1} + \int \frac{1 \mathrm{dx}}{{x}^{2} + 1}$

We calculate each integral separately

-intdx/x=-ln(∣x∣)

Let $u = {x}^{2} + 1$, $\implies$, $\mathrm{du} = 2 x \mathrm{dx}$

$\int \frac{x \mathrm{dx}}{{x}^{2} + 1} = \int \frac{2 \mathrm{du}}{u} = \frac{1}{2} \ln u$

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

Let $x = \tan \theta$, $\implies$, $\mathrm{dx} = {\sec}^{2} \theta d \theta$

and ${x}^{2} + 1 = {\tan}^{2} \theta + 1 = {\sec}^{2} \theta$

So,

$\int \frac{1 \mathrm{dx}}{{x}^{2} + 1} = \int \frac{{\sec}^{2} \theta d \theta}{\sec} ^ 2 \theta = \int d \theta = \theta$

$= \arctan x$

Putting it alltogether,

int((x-1)dx)/(x^3+x)=-ln(∣x∣)+1/2ln(x^2+1)+arctanx+C#