Answer
$\frac{3}{2(1-2x)^2}+C$
Work Step by Step
Use the substitution $u=1-2x$. Then $du=-2dx$, so $dx=-\frac{1}{2}du$.
$\int \frac{6}{(1-2x)^3}dx$
$=\int \frac{6}{u^3}*(-\frac{1}{2})du$
$=\int -\frac{3}{u^3}du$
$=\int -3u^{-3}du$
$=\frac{-3u^{-2}}{-2}+C$
$=\frac{3}{2u^2}+C$
$=\frac{3}{2(1-2x)^2}+C$