Question

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

Answer 1

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

Explanation:

First of all : `x^2 - 1 = (x-1)(x+1)`.

Partial fraction decomposition allows you to do that :

`(x+1)/(x(x^2 - 1)) = (x+1)/(x(x-1)(x+1)) = 1/(x(x-1)) = a/x + b/(x-1)` with `a,b in RR` 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.

`1/(x(x-1)) = a/x + b/(x-1) iff 1/(x-1)= a + (bx)/(x-1)`.
`x = 0 iff -1 = a`

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

So `(x+1)/(x(x^2 - 1)) = 1/(x-1) - 1/x`, which implies that `int(x+1)/(x(x^2 - 1))dx = int(1/(x-1) - 1/x)dx = intdx/(x-1) - intdx/x = lnabs(x-1) - lnabsx`