# indefinite integral

• November 8th 2007, 12:35 PM
cowboys111
indefinite integral
The original problem:
$
\int 6/(1-2x)^3 dx
$

I have got it down to this but for some reason I have no idea what to do with the 1/u^3, can someone please help?
$
-3 \int 1/u^3 du
$

thank you
• November 8th 2007, 01:18 PM
Krizalid
You almost got it!

$\frac1{u^3}=u^{-3}.$

Apply the power rule for integrals.
• November 8th 2007, 01:35 PM
cowboys111
oh wow i feel like an idiot! thanks for your time
• November 9th 2007, 02:23 PM
2taall
Did you basically get du=-2 and u=1-2x?

since you had a 6 in the numerator you pulled it out and put it in front of the integration symbol and replaced it with the -2 that you needed in that spot. Then, I guess you divided the 6 that was outside the integration symbol by
-2 to give you the -3?