Ways to prove $ \int_0^1 \frac{\ln^2(1+x)}{x}dx = \frac{\zeta(3)}{4}$?

Let us denote $I_{\pm}=\displaystyle \int_{0}^1\frac{\ln^2(1\pm x)}{x}dx$. We will express $I_+$ in terms of $I_-$, which is itself related to the standard integral representation of $\zeta(z)$ by the change of variables $x=1-e^{-t}$: $$I_-=\int_0^{\infty}\frac{t^2dt}{e^{t}-1}=2\zeta(3).$$ Indeed, we have \begin{align} \int_0^1\frac{\ln^2\frac{1+x}{1-x}}{x}dx=\int_0^{\infty}\frac{16t^2}{2\sinh 2t}dt&=\int_0^{\infty}16t^2\left(\frac{1}{e^{2t}-1}-\frac{1}{e^{4t}-1}\right)dt=\frac74 I_- \tag{1} \end{align} where the first equality is obtained by setting $x=\tanh t$. Also, it is easy to show ($x^2\to x$) that $$\int_{0}^1\frac{\ln^2(1-x^2)}{x}dx=\frac12I_-. \tag{2}$$ Summing (1) and (2), one finds that $ 2I_+ +2I_-=\left(\frac74+\frac12\right)I_-$, and hence $\displaystyle I_+=\frac{I_-}{8}=\frac{\zeta(3)}4$.

Using this answer which shows that $$ \sum_{n=1}^\infty\frac{(-1)^{n-1}}{n^2}H_n=\frac58\zeta(3) $$ and the series $$ \frac{\log(1+x)}{1+x}=\sum_{k=1}^\infty(-1)^{k-1}H_kx^k $$ we get $$ \begin{align} \int_0^1\frac{\log(1+x)^2}{x}\mathrm{d}x &=\int_0^1\log(1+x)^2\,\mathrm{d}\log(x)\\ &=-2\int_0^1\frac{\log(1+x)\log(x)}{1+x}\,\mathrm{d}x\\ &=-2\int_0^1\sum_{k=1}^\infty(-1)^{k-1}H_kx^k\log(x)\,\mathrm{d}x\\ &=2\sum_{k=1}^\infty\frac{(-1)^{k-1}H_k}{(k+1)^2}\\ &=2\sum_{k=1}^\infty\frac{(-1)^{k-1}H_{k+1}}{(k+1)^2}-2\sum_{k=1}^\infty\frac{(-1)^{k-1}}{(k+1)^3}\\ &=2\left(\frac34\zeta(3)-\sum_{k=1}^\infty\frac{(-1)^{k-1}H_k}{k^2}\right)\\ &=\frac{\zeta(3)}4 \end{align} $$