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

Dec 30, 2015

You try to split the rational function into a sum that will be really easy to integrate.

#### Explanation:

First of all : ${x}^{2} - 1 = \left(x - 1\right) \left(x + 1\right)$.

Partial fraction decomposition allows you to do that :

$\frac{x + 1}{x \left({x}^{2} - 1\right)} = \frac{x + 1}{x \left(x - 1\right) \left(x + 1\right)} = \frac{1}{x \left(x - 1\right)} = \frac{a}{x} + \frac{b}{x - 1}$ with $a , b \in \mathbb{R}$ that you have to find.

In order to find them, you have to multiply both sides by one of the polynomials at the left of the equality. I show one example to you, the other coefficient is to be found the same way.

We're gonna find $a$ : we have to multiply everything by $x$ in order to make the other coefficient disappear.

$\frac{1}{x \left(x - 1\right)} = \frac{a}{x} + \frac{b}{x - 1} \iff \frac{1}{x - 1} = a + \frac{b x}{x - 1}$.
$x = 0 \iff - 1 = a$

You do the same thing in order to find $b$ (you multiply everything by $\left(x - 1\right)$ then you choose $x = 1$), and you find out that $b = 1$.

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