sequences and series - How to evaluate $int_0^1frac{log^2(1+x)}xmathrm dx$?

The definite integral

$$\int_0^1\frac{\log^2(1+x)}x\mathrm dx=\frac{\zeta(3)}4$$

arose in my answer to this question. I couldn't find it treated anywhere online. I eventually found two ways to evaluate the integral, and I'm posting them as answers, but they both seem like a complicated detour for a simple result, so I'm posting this question not only to record my answers but also to ask whether there's a more elegant derivation of the result.

Note that either using the method described in this blog post or substituting the power series for $\log(1+x)$ and using

$$\frac1k\frac1{s-k}=\frac1s\left(\frac1k+\frac1{s-k}\right)\;$$

yields

$$
\int_0^1\frac{\log^2(1+x)}x\mathrm dx=2\sum_{n=1}^\infty\frac{(-1)^{n+1}H_n}{(n+1)^2}\;.
$$

However, since the corresponding identity without the alternating sign is used to obtain the sum by evaluating the integral and not vice versa, I'm not sure that this constitutes progress.