Prove that $\int_0^{\infty}\left(\frac{\log (1+x)}{x}\right)^2dx$ converges. Prove that $\int_0^{\infty}\left(\frac{\log (1+x)}{x}\right)^2dx$ converges.
I'm not sure where to begin here, perhaps show that it is bounded by something? This is an exercise I'm doing to review calculus.
 A: There is potential trouble near $0$, and also for large $x$. It is useful to break up the integral into the integral from $0$ to (say) $1$, and from $1$ to $\infty$. If each integral converges, our integral does.
From $0$ to $1$: We look at the behaviour of $\left(\frac{\ln(1+x)}{x}\right)^2$ for positive $x$ near $x=0$. 
By L'Hospital's Rule, we have $\lim_{t\to 0}\frac{\ln(1+x)}{x}=1$. So, appearances to the contrary, our function is very well behaved near $x=0$. If we define it to be $0$ at $x=0$, it is continuous. Thus the integral from $0$ to $1$ converges.
From $1$ to $\infty$: Rewrite our function as $\frac{1}{x^{3/2}}\frac{\ln(1+x)}{x^{1/2}}$.
By using L'Hospital's Rule, we can show that $\lim_{x\to\infty}\frac{\ln(1+x)}{x^{1/2}} =0$.
Thus for large enough $x$, $\frac{\ln(1+x)}{x^{1/2}}\le 1$. It follows that for large enough $x$, our function is $\lt \frac{1}{x^{3/2}}$. By Comparison, the integral from $1$ to $\infty$ converges.
A: The convergence has been established by @André Nicolas. 
By integrating by parts, one may observe there is a closed form of the given integral, 
$$
\begin{align}
\int_0^\infty \left(\frac{\ln (1+x)}{x} \right)^2dx&=\left.- \frac1x \cdot \ln^2(1+x) \right]_0^\infty+2\int_0^\infty \frac{\ln(1+x)}{x(1+x)}\:dx
\\&=\color{red}{0}+2 \cdot \frac {\pi^2}6
\\&=\color{blue}{\frac {\pi^2}3}
\end{align}
$$ where we have used
$$
2\int_0^\infty \frac{\ln(1+x)}{x(1+x)}\:dx=4\int_0^\infty \frac{\ln(1+t^2)}{t(1+t^2)}\:dt=4\cdot \frac {\pi^2}{12}=\color{blue}{\frac {\pi^2}3}
$$ proved here.
