How do you integrate [(x^2 + 1)/(x^2 - 1)]dx[x2+1x2−1]dx?

1 Answer
Shwetank Mauria
Jun 20, 2016

int[(x^2+1)/(x^2-1)]dx=x-ln(x+1)+ln(x-1)+c∫[x2+1x2−1]dx=x−ln(x+1)+ln(x−1)+c

Explanation:

int[(x^2+1)/(x^2-1)]dx∫[x2+1x2−1]dx

= int[(x^2-1+2)/(x^2-1)]dx∫[x2−1+2x2−1]dx

= int[1+2/(x^2-1)]dx∫[1+2x2−1]dx

= x+int2/(x^2-1)dxx+∫2x2−1dx

As (x^2-1)=(x+1)(x-1)(x2−1)=(x+1)(x−1), let 2/((x+1)(x-1))hArrA/(x+1)+B/(x-1)2(x+1)(x−1)⇔Ax+1+Bx−1 i.e.

2/((x+1)(x-1))hArr(A(x-1)+B(x+1))/((x+1)(x-1))2(x+1)(x−1)⇔A(x−1)+B(x+1)(x+1)(x−1) or

2/((x+1)(x-1))hArr((A+B)x-A+B)/((x+1)(x-1))2(x+1)(x−1)⇔(A+B)x−A+B(x+1)(x−1)

i.e. A+B=0A+B=0 and -A+B=2−A+B=2 and adding them we get 2B=22B=2

or B=1B=1 and A=-1A=−1. Hence

int[(x^2+1)/(x^2-1)]dx=x+int[-1/(x+1)+1/(x-1)]dx∫[x2+1x2−1]dx=x+∫[−1x+1+1x−1]dx

= x-ln(x+1)+ln(x-1)+cx−ln(x+1)+ln(x−1)+c

Related questions
See all questions in Integrals of Rational Functions
Impact of this question
60384 views around the world
You can reuse this answer
Creative Commons License