Answer
Converges to $\ln 2$
Work Step by Step
Type I.
$I=\displaystyle \int_{1}^{\infty}\frac{1}{x^{2}+x}dx=\lim_{t\rightarrow\infty}\int_{1}^{t}\frac{1}{x(x+1)}dx$
$\displaystyle \frac{1}{x(x+1)}=\frac{A}{x}+\frac{B}{x+1}$
$\displaystyle \frac{1}{x(x+1)}=\frac{A(x+1)+Bx}{x}$
$\displaystyle \frac{1}{x(x+1)}=\frac{(A+B)x+A}{x}\Rightarrow\left\{\begin{array}{ll}
A=1 & \\
A+B=0 & \Rightarrow B=-1
\end{array}\right\}$
$I=\displaystyle \lim_{t\rightarrow\infty}\int_{1}^{t}(\frac{1}{x}-\frac{1}{x+1})dx\quad $
$=\displaystyle \lim_{t\rightarrow\infty}[\ln|x|-\ln|x+1|]_{1}^{t}$
$=\displaystyle \lim_{t\rightarrow\infty}\left[\ln\left|\frac{x}{x+1}\right|\right]_{1}^{t}$
$=\displaystyle \lim_{t\rightarrow\infty}(\ln\frac{t}{t+1}-\ln\frac{1}{2})$
$=0-\displaystyle \ln\frac{1}{2}$
$=\ln 2$
(converges)