Answer
Divergent
Work Step by Step
Let \[I=\int_{-1}^{2}\frac{x}{(x+1)^2}dx\]
Since $-1$ is point of infinite discontinuity of integrand $\large\frac{x}{(x+1)^2}$
\[I=\lim_{t\rightarrow -1^+}\int_{t}^{2}\frac{x}{(x+1)^2}dx\;\;\;\ldots (1)\]
Let \[I_1=\int\frac{x}{(x+1)^2}dx\]
Substitute $\; r=x+1\;\;\;\ldots (2)$
\[\Rightarrow dr=dx\]
\[I_1=\int\frac{r-1}{r^2}dr\]
\[I=\int\left(\frac{1}{r}-r^{-2}\right)dr\]
\[I=\ln r-\frac{r^{-1}}{-1}=\ln r+\frac{1}{r}\]
From (2)
\[I_1=\ln (x+1)+\frac{1}{(x+1)}\;\;\;\ldots (3)\]
Using (3) in (1)
\[I=\lim_{t\rightarrow -1^+}\left[\ln (x+1)+\frac{1}{(x+1)}\right]_{t}^{2}\]
\[I=\lim_{t\rightarrow -1^+}\left[\ln 3+\frac{1}{3}-\ln (t+1)-\frac{1}{(t+1)}\right]\]
$\;\;\;\;\;\;\;\;\;\;\;\;= $ does not exist
Since limit on R.H.S. of (1) does not exists so $I$ is divergent.
