Question

How do you find the integral of `(u-4)/ [u^2 +1]`?

Answer 1

`1/2In abs(u^2+1)-4tan^(-1)(u^2+1)+c`

Explanation:

`int(u-4)/(u^2+1) du`
=`intu/(u^2+1)-(4)/(u^2+1) du`
=`1/2In abs(u^2+1)-4tan^(-1)(u^2+1)+c`