# Where am i going wrong on this simple integration

• March 28th 2010, 04:50 PM
vinson24
Where am i going wrong on this simple integration
$\int\ (4-2x)^3$
First i factor out a -2
$-2\int\ (x-2)^3$
then $u=x-2$
$du=1$
$-2\int\ u^3du$
$\frac{-1}{2}(x-2)^4+C$
but the book has $-2(x-2)^4+C$
• March 28th 2010, 04:54 PM
pickslides
$
(4-2x)^3\neq-2\ (x-2)^3
$
• March 28th 2010, 05:17 PM
skeeter
let $u = 4-2x$

$du = -2 \, dx$