## Calculus 10th Edition

$\ln|x-1|+\frac{1}{2(x-1)^2}+C$
let u=x-1, du=dx $u=x-1$ $u+1=x$ $u-1=x-1-1=x-2$ $\int\frac{x(x-2)}{(x-1)^3}dx$ $=\int\frac{(u+1)(u-1)}{u^3}du$ $=\int\frac{u^2-1}{u^3}du$ $=\int(u^2-1)u^{-3}du$ $=\int(u^{-1}-u^{-3})du$ $=ln|u|+\frac{1}{2}u^{-2}+C$ $=ln|x-1|+\frac{1}{2(x-1)^2}+C$