The original problem:

$\displaystyle

\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?

$\displaystyle

-3 \int 1/u^3 du

$

thank you

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- Nov 8th 2007, 12:35 PMcowboys111indefinite integral
The original problem:

$\displaystyle

\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?

$\displaystyle

-3 \int 1/u^3 du

$

thank you - Nov 8th 2007, 01:18 PMKrizalid
You almost got it!

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

Apply the power rule for integrals. - Nov 8th 2007, 01:35 PMcowboys111
oh wow i feel like an idiot! thanks for your time

- Nov 9th 2007, 02:23 PM2taall
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?