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

1 Answer
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#