Answer
$\ln |x-1|-\dfrac{1}{x-1}+C$
Work Step by Step
We will solve the given integral by using u-substitution method.
Let us consider that $u=x-1 \implies dx=du$
$\displaystyle \int \dfrac{x}{(x-1)^2} \ dx=\int \displaystyle \dfrac{u+1}{u^2} \ du$
or, $=\int \displaystyle (\dfrac{1}{u}+u^{-2}) \ du$
or, $= \ln |u|-\dfrac{1}{u}+C$
Now, we will use back substitution $u=x-1$
Therefore, we have:
$\displaystyle \int \dfrac{x}{(x-1)^2} \ dx=\ln |x-1|-\dfrac{1}{x-1}+C$
