## Calculus 10th Edition

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