Hi, I posted this in pre-calculus section accidentally (as I got a bit confused):

$\displaystyle \int \frac{y\,(y-8)}{(y-4)^2} dy $

using the substitution u = y - 4

The answer is

$\displaystyle \frac{y^2-8y+32}{y-4}+c$

But I keep on getting

$\displaystyle \frac{y^2-4y+16}{y-4}+c$

Any help would be greatly appreciated