Question

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

Answer 1

`int x/(x+1) dx = x-ln abs(x+1)+C`

Explanation:

`int x/(x+1) dx`

`=int (x+1-1)/(x+1) dx`

`=int (1-1/(x+1)) dx`

`= x-ln abs(x+1)+C`