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

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.

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

$$ \frac {{u}^{2}{{\rm e}^{u}}}{ \left( {{\rm e}^{u }}-1 \right) ^{2}}\sim_{u\sim \infty} \frac {{u}^{2}{{\rm e}^{u}}}{ ({{\rm e}^{u }}) ^{2}}=\dots\,. $$

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
$(6)$ $\eta (2)=(1-2^{1-2})\zeta(2)=\frac{\pi^2}{12}$

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: $$\int_{0}^{+\infty}\frac{\log^2(1+x)}{x^2}\,dx = 2\sum_{n=1}^{+\infty}\frac{1}{n^2}=\frac{\pi^2}{3}.$$