Question

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

Answer 1

`int((x^2-1)/(x+1)) = (x^2)/2 -x+c`

Explanation:

We know that `x^2 - 1 = (x+1)(x-1)`, by the difference of squares so we can rewrite the integral to be

`int((x+1)(x-1))/(x+1)dx = int(x-1)dx = intxdx - intdx = x^2/2 - x + c`

Or if you prefer

`int((x^2-1)/(x+1)) = (x^2-2x+c)/2`