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

1 Answer
Topscooter
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 = (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

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