Answer
\[\ln|x-1|-3(x-1)^{-1} -\frac{3}{2}(x-1)^{-2}-\frac{1}{3}(x-1)^{-3}+C\]
Work Step by Step
\[I=\int x^3(x-1)^{-4}dx \;\;\;\;\ldots(1)\]
\[I=\int\frac{x^3}{(x-1)^4}dx\]
Substitute $x-1=t \;\;\;\;\ldots (2)$
\[\Rightarrow dx=dt\]
\[I=\int\frac{(t+1)^3}{t^4}dt\]
\[I=\int\frac{t^3+1+3t^2+3t}{t^4}dt\]
\[I=\int\left[\frac{1}{t}+t^{-4}+3t^{-2}+3t^{-3}\right]dt\]
\[I=\ln|t| -\frac{1}{3}(t)^{-3} -3(t)^{-1} -\frac{3}{2}(t)^{-2}+C\]
Using (2)
\[I=\ln|x-1| -\frac{1}{3}(x-1)^{-3} -3(x-1)^{-1} -\frac{3}{2}(x-1)^{-2}+C\]
\[I=\ln|x-1|-3(x-1)^{-1} -\frac{3}{2}(x-1)^{-2}-\frac{1}{3}(x-1)^{-3}+C\]
Hence \[I=\ln|x-1|-3(x-1)^{-1} -\frac{3}{2}(x-1)^{-2}-\frac{1}{3}(x-1)^{-3}+C.\]
