Question

How do you integrate `int( 2 / (x + 1) )^2dx`?

Answer 1

`-4/(x + 1) + C`

Explanation:

Let `u = 2/(x + 1)` and `du = -2/(x + 1)^2dx` and `dx = -(x + 1)^2/2du`

`=> int(2/(x + 1))^2 * -(x + 1)^2/2 du`

`=> int(4/(x + 1)^2 * -(x + 1)^2/2) du`

`=> int(-2)du`

Use the rule `int(x^n)dx = x^(n + 1)/(n + 1) + C`

`=> -2u`

`=> -2(2/(x + 1)) + C`

`=> -4/(x + 1) + C`

Hopefully this helps!