# Math Help - Integration by substitution

1. ## Integration by substitution

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

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

using the substitution u = y - 4

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

But I keep on getting
$\frac{y^2-4y+16}{y-4}+c$

Any help would be greatly appreciated

2. ## Re: Integration by substitution

$\int \frac{(u+4)\,(u-4)}{u^2} du$
$=\int \frac{(u^2-16)}{u^2} du$
$=\int 1-\frac{16}{u^2} du$
$=u +\frac{16}{u}+c$
$=y-4 +\frac{16}{y-4}+c$
$=\frac{y^2-8y+32}{y-4}+c$

3. ## Re: Integration by substitution

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