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
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?