Answer
$$\ln \frac{3}{2}-\frac{1}{6} $$
Work Step by Step
Given $$\int_{1}^{2} \frac{x}{(x+1)^{2}} d x$$
Let $u=x+1\ \ \Rightarrow \ \ du=dx$ and at $ x=1\to u=2$, at $ x=2\to u=3$
\begin{align*}
\int_{1}^{2} \frac{x}{(x+1)^{2}}d x&=\int_{2}^{3} \frac{ u-1 }{u^2} d u\\
&=\int_{2}^{3} \left( \frac{1 }{u}-u^{-2} \right)du\\
&= \ln u+\frac{1}{u}\bigg|_{2}^3\\
&=\ln \frac{3}{2} + \frac{1}{3}-\frac{1}{2} \\
&=\ln \frac{3}{2}-\frac{1}{6}
\end{align*}
