Integration by substitution

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 :)

Re: Integration by substitution

$\displaystyle \int \frac{(u+4)\,(u-4)}{u^2} du $

$\displaystyle =\int \frac{(u^2-16)}{u^2} du $

$\displaystyle =\int 1-\frac{16}{u^2} du $

$\displaystyle =u +\frac{16}{u}+c $

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

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

Re: Integration by substitution

Thank you so much, your working is really clear and helpful, I know exactly where I went wrong now! :)