Answer
$\displaystyle \ln\frac{3}{2}-\frac{1}{6}$
Work Step by Step
$I=\displaystyle \int_{1}^{2}\frac{x}{(x+1)^{2}}dx= \left[\begin{array}{ll}
t=x+1 & \\
dt=dx & x\in[1,2]\rightarrow t\in[2,3]
\end{array}\right]$
$=\displaystyle \int_{2}^{3}\frac{t-1}{t^{2}}dt=$
$=\displaystyle \int_{2}^{3}(\frac{1}{t}-\frac{1}{t^{2}})dt$
$=\left[\ln|x|-\frac{t^{-1}}{-1}\right]_{2}^{3}$
$=\left[\ln|x|+\frac{1}{t}\right]_{2}^{3}$
$=(\displaystyle \ln 3+\frac{1}{3})-(\ln 2+\frac{1}{2})$
$=\displaystyle \ln\frac{3}{2}-\frac{1}{6}$
