Answer
\[\frac{1}{4}x^4+\frac{1}{3}x^3+\frac{1}{2}x^2+x+\ln|x-1|+C\]
Work Step by Step
Let \[I=\int\frac{x^4}{x-1}dx\]
\[I=\int\frac{(x-1)(x^3+x^2+x+1)+1}{x-1}dx\]
\[I=\int\left[x^3+x^2+x+1+\frac{1}{x-1}\right]dx\]
\[I=\frac{1}{4}x^4+\frac{1}{3}x^3+\frac{1}{2}x^2+x+\ln|x-1|+C\]
Hence \[I=\frac{1}{4}x^4+\frac{1}{3}x^3+\frac{1}{2}x^2+x+\ln|x-1|+C\].