Answer
Convergent $\;,\;$ $\ln 2$
Work Step by Step
Let \[I=\int_{1}^{\infty}\frac{1}{x^2+x}dx\]
\[I=\lim_{t\rightarrow \infty}\int_{1}^{t}\frac{1}{x^2+x}dx\;\;\;\ldots(1)\]
Let \[I_1=\int\frac{1}{x^2+x}dx\]
\[I_1=\int\frac{1}{x(x+1)}dx\]
\[I_1=\int\left[\frac{1}{x}-\frac{1}{x+1}\right]dx\]
\[I_1=\ln|x|-\ln |x+1|\]
\[I_1=\ln\left|\frac{x}{x+1}\right|\;\;\;\ldots (2)\]
Using (2) in (1)
\[I=\lim_{t\rightarrow \infty}\left[\ln\left|\frac{x}{x+1}\right|\right]_{1}^{t}\]
\[I=\lim_{t\rightarrow \infty}\left[\ln\left|\frac{t}{t+1}\right|-\ln\left|\frac{1}{2}\right|\right]\]
\[I=\lim_{t\rightarrow \infty}\left[\ln\left|\frac{1}{1+\frac{1}{t}}\right|-\ln\left|\frac{1}{2}\right|\right]\]
\[I=-\ln\frac{1}{2}=\ln 2\]
Since limit on R.H.S. of (1) exists so $I$ is convergent
and $I=\ln 2$.
