Answer
$\displaystyle \ln|x-1|-3(x-1)^{-1}-\frac{3}{2}(x-1)^{-2}-\frac{1}{3}(x-1)^{-3}+C$
Work Step by Step
$I=\displaystyle \int x^{3}(x-1)^{-4}dx$
We can't distribute as it stands, but with
sub: $\left[\begin{array}{ll}
u=x-1 & x=u+1\\
du=dx &
\end{array}\right]$, the integral becomes $\displaystyle \int(u+1)^{3}u^{-4}du$.
Expand the cube and distribute:
$=\displaystyle \int(u+1)^{3}u^{-4}du=\int(u^{3}+3u^{2}+3u+1)u^{-4}du$
$=\displaystyle \int(u^{-1}+3u^{-2}+3u^{-3}+u^{-4})du$
... these are table integrals of type 1 and 2:
$=\displaystyle \ln|u|-3u^{-1}-\frac{3}{2}u^{-2}-\frac{1}{3}u^{-3}+C$
... bring back x
$=\displaystyle \ln|x-1|-3(x-1)^{-1}-\frac{3}{2}(x-1)^{-2}-\frac{1}{3}(x-1)^{-3}+C$
