I'd like to have my solution verified for this one. I'd like to show that $$\int\limits_{1}^{\infty}\dfrac{1}{x^2+x}\text{ d}x$$ is convergent. Notice, by partial fraction decomposition, that $$\dfrac{1}{x^2+x} = \dfrac{1}{x}-\dfrac{1}{x+1}\text{.}$$ An antiderivative of $\dfrac{1}{x^2+x}$ can be easily seen to be $\ln|x|-\ln|x+1| = \ln\left(\left|\dfrac{x}{x+1}\right|\right)$.
As $x \to \infty$, $\ln\left(\left|\dfrac{x}{x+1}\right|\right) \to \ln(1) = 0$ by continuity of $\ln$.
Thus $$\int\limits_{1}^{\infty}\dfrac{1}{x^2+x}\text{ d}x = 0 - \ln\left(\dfrac{1}{2}\right) = \ln\left[\left(\dfrac{1}{2}\right)^{-1}\right] = \ln(2)\text{.}$$
Your solution is OK. Observe that $$ 0<\int\limits_{1}^{\infty}\dfrac{1}{x^2+x}\text{ d}x<\int\limits_{1}^{\infty}\dfrac{1}{x^2}\text{ d}x=\left[ -\frac{1}{x}\right]_1^\infty=1. $$