Answer
Diverges.
Work Step by Step
The integrand has a discontinuity at $-1$, so this is a type 2.b improper integral.
$\displaystyle \int_{a}^{b}f(x)dx=\lim_{t\rightarrow a^{+}}\int_{t}^{b}f(x)dx$
$\displaystyle \int_{-1}^{2}\frac{x}{(x+1)^{2}}dx=\lim_{t\rightarrow-1^{+}}\int_{t}^{2}\frac{x}{(x+1)^{2}}dx$
$\left[\begin{array}{ll}
u=x+1 & \\
du=dx & \\
x=-1\Rightarrow u=0, & x=2\Rightarrow u=3
\end{array}\right]$
$=\displaystyle \lim_{u\rightarrow 0^{+}}\int_{u}^{3}\frac{u-1}{u^{2}}dx$
$=\displaystyle \lim_{u\rightarrow 0^{+}}\left[\ln|u|-\frac{u^{-1}}{-1}\right]_{u}^{3}$
$=\displaystyle \lim_{u\rightarrow 0^{+}}\left[\ln|3|+\frac{1}{3}\right]-\lim_{u\rightarrow 0^{+}}\left[\ln|u|+\frac{1}{u}\right]$
Diverges, because
$\displaystyle \lim_{u\rightarrow 0^{+}}\left[\ln|u|+\frac{1}{u}\right]=\lim_{u\rightarrow 0^{+}}\left[(u\ln u+1)\frac{1}{u}\right]$
Now,
$\displaystyle \lim_{u\rightarrow 0^{+}}u\ln u =\displaystyle \lim_{u\rightarrow 0^{+}}\frac{\ln u}{\frac{1}{u}}=$(L,Hosp)$=\displaystyle \lim_{u\rightarrow 0^{+}}\frac{1}{u(-u^{-2})}=\lim_{u\rightarrow 0^{+}}(-u)=0$, so
$\displaystyle \lim_{u\rightarrow 0^{+}}\left[\ln|u|+\frac{1}{u}\right]=(0+1)\lim_{u\rightarrow 0^{+}}\frac{1}{u}=\infty$
