# Integration by substitution

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• Dec 1st 2011, 11:45 AM
sakuraxkisu
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

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

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

Any help would be greatly appreciated :)
• Dec 1st 2011, 12:03 PM
wnvl
Re: Integration by substitution
$\int \frac{(u+4)\,(u-4)}{u^2} du$
$=\int \frac{(u^2-16)}{u^2} du$
$=\int 1-\frac{16}{u^2} du$
$=u +\frac{16}{u}+c$
$=y-4 +\frac{16}{y-4}+c$
$=\frac{y^2-8y+32}{y-4}+c$
• Dec 1st 2011, 12:07 PM
sakuraxkisu
Re: Integration by substitution
Thank you so much, your working is really clear and helpful, I know exactly where I went wrong now! :)