Lời giải:
Đặt $x-1=t$ thì:
\(\int \frac{x^2}{(x-1)^5}dx=\int \frac{(t+1)^2}{t^5}d(t+1)\\ =\int \frac{t^2+2t+1}{t^5}dt\\ =\int t^{-3}dt+2\int t^{-4}dt+\int t^{-5}dt\\ =\frac{t^{-2}}{-2}+2.\frac{t^{-3}}{-3}+\frac{t^{-4}}{-4}+C\\ =\frac{1}{2(x-1)^2}-\frac{2}{3(x-1)^3}-\frac{1}{4(x-1)^4}+C\)