Question

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

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

Answer 1

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.