Calculate the integral or prove that it diverges

Question

Given integral

$$\int_{1}^{+ \infty} \frac{\ln(x^2 + x)}{x}\, dx,$$

calculate it or prove that it diverges.

I've tried to use integration by parts, but after second application I got the initial integral.

I also thought about the following: since

$$\frac{\ln(x^2 + x)}{x} \geqslant 0 \ \forall x \in [1, +\infty),$$

I may want to get an lower bound (some $g(x)$) such that

$$\frac{\ln(x^2 + x)}{x} \geqslant g(x) \forall x \in [1, +\infty)$$

and then prove that $g$ diverges. Unfortunately I have no idea, which function to take as $g$.

Answer 1

The function $\ln(x^2+x)$ is increasing on $[1,\infty)$, hence $$ \frac{\ln(x^2+x)}{x}\geq \frac{\ln(2)}{x} $$ for all $x\geq 1$. And since $$ \int_1^{\infty}\frac{dx}{x}=\infty $$ this implies that $$ \int_{1}^{\infty}\frac{\ln(x^2+x)}{x}=\infty$$ as well.

Answer 2

$$\begin{array}{rcl} \displaystyle \int_1^\infty \frac{\ln(x^2 + x)}{x}\ \mathrm dx &>&\displaystyle \int_1^\infty \frac{\ln(x^2)}{x}\ \mathrm dx \\ &=&\displaystyle 2\int_1^\infty \frac{\ln x}{x}\ \mathrm dx \\ &=&\displaystyle 2\int_1^\infty \ln x \ \mathrm d(\ln x) \\ &=&\displaystyle 2\left(\dfrac12 (\ln x)^2\right)_1^\infty \\ &=&\displaystyle \lim_{n\to\infty} (\ln n)^2 \\ &=&\infty \end{array}$$