Answer
$-\dfrac{1}{(x-2)}-\dfrac{1}{(x-2)^2} +C$
Work Step by Step
We will solve the given integral by using u-substitution method.
Let us consider that $u=x-2 \implies dx=du$
$\displaystyle \int \dfrac{x}{(x-2)^3} \ dx=\int \displaystyle \dfrac{u+2}{u^3} \ du$
or, $=\int \displaystyle (u^{-2}+2u^{-3}) \ du$
or, $= \dfrac{u^{-1}}{-1}+\dfrac{2u^{-2}}{-2} +C$
Now, we will use back substitution $u=x-2$
Therefore, we have:
$\displaystyle \int \dfrac{x}{(x-2)^3} \ dx=-\dfrac{1}{(x-2)}-\dfrac{1}{(x-2)^2} +C$
