Show that $\int^\infty_0\left(\frac{\ln(1+x)} x\right)^2dx$ converge. 
Show that $$\int\limits^\infty_0\left(\frac{\ln(1+x)} x\right)^2dx$$ converge.

I have utterly no clue on this integral. Please give me some hints. Thanks you.
 A: Since 
$$\lim_{x\to0}\left(\frac{\ln(1+x)}{x}\right)^2=1$$
then the integral
$$\int_0^1\left(\frac{\ln(1+x)}{x}\right)^2dx$$
exists, moreover we have
$$\ln^2(1+x)=_\infty o(x^{1/2})$$
so
$$\left(\frac{\ln(1+x)}{x}\right)^2=_\infty o\left(\frac{1}{x^{3/2}}\right)$$
and then the integral
$$\int_1^\infty \left(\frac{\ln(1+x)}{x}\right)^2dx$$
also exists. Conclude.
A: If you use the change of variables $\ln(1+x)=u$ things will be easier. The integral becomes
$$ \int _{0}^{\infty }\!{\frac {{u}^{2}{{\rm e}^{u}}}{ \left( {{\rm e}^{u
}}-1 \right) ^{2}}}{du}.$$
Now you can see that close to $0$ the integrand behaves as 
$$ \frac {{u}^{2}{{\rm e}^{u}}}{ \left( {{\rm e}^{u
}}-1 \right) ^{2}}\sim_{u\sim 0}  \frac {{u}^{2}{{\rm e}^{0}}}{ \left( (1+u)-1 \right) ^{2}} = 1 $$
which is integrable. Note that we used Taylor series the function

$$ e^{u} = 1+u+\frac{u^2}{2!}+\dots\,.  $$

Try to do the same with the other end of the interval and to compare the integrand with an integrable function.
Added: For the other end you should be able to see that
$$  \frac {{u}^{2}{{\rm e}^{u}}}{ \left( {{\rm e}^{u
}}-1 \right) ^{2}}\sim_{u\sim \infty} \frac {{u}^{2}{{\rm e}^{u}}}{  ({{\rm e}^{u
}})
^{2}}=\dots\,. $$
A: If you don't mind, I would like to evaluate this integral.
\begin{align}
\int^\infty_0\frac{\ln^2(1+x)}{x^2}dx\tag1
&=2\int^\infty_0\frac{\ln(1+x)}{x(1+x)}dx\\\tag2
&=2\left[\int^1_0\frac{\ln(1+x)}{x(1+x)}dx+\int^1_0\frac{\ln\left(1+\frac{1}{x}\right)}{1+x}dx\right]\\\tag3
&=2\left[\int^1_0\frac{\ln(1+x)}{x}dx-\int^1_0\frac{\ln x}{1+x}dx\right]\\\tag4
&=2\left[-\mathrm{Li}_2(-1)-\sum_{n \ge 0}(-1)^n\int^1_0x^n\ln{x}dx\right]\\\tag5
&=2\left[\frac{\pi^2}{12}-\underbrace{\sum_{n \ge 1}\frac{(-1)^n}{n^2}}_{-\eta \ (2)}\right]\\\tag6
&=2\left(\frac{\pi^2}{12}+\frac{\pi^2}{12}\right)\\
&=\frac{\pi^2}{3}
\end{align}
It automatically follows that the integral converges. 
Explanation:
$(1)$ Integrate by parts
$(2)$ Split the integral into $2$ and substitute $x \mapsto \frac{1}{x}$ for the second integral
$(3)$ Simplify using partial fractions and properties of logarithms
$(4)$ Expand $\frac{1}{1+x}$ as a series
$(5)$ Integrate by parts
$(6)$ $\eta (2)=(1-2^{1-2})\zeta(2)=\frac{\pi^2}{12}$
A: The integrand function is integrable over the $I=[0,1]$ interval since it is continuous over there.
Moreover, for any $x\geq 1$ we have $\log(1+x)\leq x^{3/7}$, hence the integrand function is integrable over $[1,+\infty)$ since it is positive and bounded by the integrable function $\frac{1}{x^{8/7}}$.
By exploiting the change of variable suggested by Mhenni Benghorbal, an integration by parts and a geoemtric series, you can also check that:
$$\int_{0}^{+\infty}\frac{\log^2(1+x)}{x^2}\,dx = 2\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{3}.$$ 
A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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$\ds{\int_{0}^{\infty}{\ln^{2}\pars{1 + x} \over x^{2}}\,\dd x:\ {\large ?}}$.

\begin{align}&\color{#c00000}{\int_{0}^{\infty}%
{\ln^{2}\pars{1 + x} \over x^{2}}\,\dd x}
=\int_{0}^{\infty}{1 \over x}\,2\ln\pars{1 + x}\,{1 \over 1 + x}\,\dd x
\\[3mm]&=2\int_{0}^{\infty}\bracks{%
{\ln\pars{1 + x} \over x} - {\ln\pars{1 + x} \over x + 1}}\,\dd x
\end{align}

With $\ds{\Lambda > 0}$:
\begin{align}\int_{0}^{\Lambda}{\ln\pars{1 + x} \over x}\,\dd x
&=\int_{0}^{-\Lambda}{\ln\pars{1 - x} \over x}\,\dd x
=-\int_{0}^{-\Lambda}{\rm Li}_{2}'\pars{x}\,\dd x
=-{\rm Li}_{2}\pars{-\Lambda}
\\[3mm]&={\rm Li}_{2}\pars{-\,{1 \over \Lambda}} + {\pi^{2} \over 6}
+\half\,\ln^{2}\pars{\Lambda}
\\[5mm]\int_{0}^{\Lambda}{\ln\pars{1 + x} \over x + 1}\,\dd x
&=\half\,\ln^{2}\pars{1 + \Lambda}
\end{align}

Then,
  \begin{align}\color{#c00000}{\int_{0}^{\Lambda}%
{\ln^{2}\pars{1 + x} \over x^{2}}\,\dd x}
={\pi^{2} \over 3} +
2\,{\rm Li}_{2}\pars{-\,{1 \over \Lambda}}
-\bracks{\ln^{2}\pars{1 + \Lambda} - \ln^{2}\pars{\Lambda}} 
\end{align}

Note that
$\ds{\lim_{\Lambda \to \infty}{\rm Li}_{2}\pars{-\,{1 \over \Lambda}} = 0}$ and
$$
\ln^{2}\pars{1 + \Lambda} - \ln^{2}\pars{\Lambda}
=\bracks{\Lambda\ln\pars{1 + {1 \over \Lambda}}}\,
{\ln\pars{1 + \Lambda} + \ln\pars{\Lambda} \over \Lambda}\
\stackrel{\Lambda \to \infty}{\Huge\longrightarrow}\ {0\atop}
$$

such that
  $$
\color{#66f}{\large\int_{0}^{\infty}%
{\ln^{2}\pars{1 + x} \over x^{2}}\,\dd x = {\pi^{2} \over 3}} \approx {\tt 3.2899}
$$

