How do you integrate (2x)/((x-1)(x+1)) using partial fractions?

1 Answer
Jul 18, 2016

ln|x+1|+ln|x-1|+C where C is a constant

Explanation:

The given expression can be written as partial sum of fractions:

(2x)/((x+1)(x-1))=1/(x+1)+1/(x-1)

Now let's integrate :

int(2x)/((x+1)(x-1))dx

int1/(x+1)+1/(x-1)dx

int1/(x+1)dx+int1/(x-1)dx

int(d(x+1))/(x+1)+int(d(x-1))/(x-1)

ln|x+1|+ln|x-1|+C where C is a constant