How to solve this integral int 3/(x+1)^2 sqrt((x-1)/(x+1)) \ dx?

3/(x+1)^2sqrt((x-1)/(x+1))

1 Answer
Jun 12, 2018

Please see below.

Explanation:

Seeing (x+1)^2 in the denominator of one factor and the other involving (x-1)/(x+1) it makes sense to consider the possibility of using a substitution. This is suggested by the fact that the derivative of (x-1)/(x+1) has denominator (x+1)^2.

Let u = (x-1)/(x+1), then du = (1(x+1)-(x-1)(1))/(x+1)^2 dx = 2/(x+1)^2 dx

int 3/(x+1)^2 sqrt((x-1)/(x+1)) \ dx = 3/2 int underbrace(((x-1)/(x+1))^(1/2))_(u^(1/2)) underbrace(2/(x+1)^2 dx)_(du)

= 3/2int u^(1/2) du = 3/2 2/3 u^(3/2) +C = u^(3/2) +C

= ((x-1)/(x+1))^(3/2) +C

Check this answer by differentiating to make sure we get the original integrand.