<s>[INST] Ways to prove $ \int_0^1 \frac{\ln^2(1+x)}{x}dx = \frac{\zeta(3)}{4}$? I am wondering if we can show in a simple way that
$$
I=\int_0^1 \frac{\ln^2(1+x)}{x}dx = \int_1^2 \frac{\ln^2(t)}{t-1}dt = \frac{\zeta(3)}{4}.
$$
Because the end result is very simple, I suspect that there might be a fast way to prove this.
Can you prove it without using polylog identities? Complex analysis is allowed.
It may be easier to show the equivalent identity
$$
\sum_{k=1}^\infty \frac{(-1)^k H_k}{k^2} = -\frac{5 \zeta (3)}{8}
$$
I know you can do that one with the generating function of the harmonic numbers, but that gives a nasty expression with polylogs which I would like to avoid.
 [/INST] Here is a particularly efficient way to get to your Euler sum. 
In this post I show that
$$\ln^2 (1 - x) = 2 \sum_{n = 2}^\infty \frac{H_{n - 1} x^n}{n}.$$
Replacing $x $ with $-x$ gives
$$\ln^2 (1 + x) = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1} x^n}{n}.$$
So if we replace the term $\ln^2 (1 + x)$ with its above Maclaurin series expansion the integral becomes
$$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1}}{n} \int_0^1 x^{n - 1} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_{n - 1}}{n^2}.$$
Making use of the following property for harmonic numbers
$$H_n = H_{n - 1} + \frac{1}{n},$$
the integral can be expressed as
$$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = 2 \sum_{n = 2}^\infty \frac{(-1)^n H_n}{n^2} - 2 \sum_{n = 2}^\infty \frac{(-1)^n}{n^3} = 2 \sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} - 2 \sum_{n = 1}^\infty \frac{(-1)^n}{n^3}.$$
For the sums, as you note
$$\sum_{n = 1}^\infty \frac{(-1)^n H_n}{n^2} = -\frac{5}{8} \zeta (3),$$
and
$$\sum_{n = 1}^\infty \frac{(-1)^n}{n^3} = - \sum_{n = 1}^\infty \frac{(-1)^{n - 1}}{n^3} = - \eta (3) = -(1 - 2^{1-3}) \zeta (3) = -\frac{3}{4} \zeta (3),$$
where $\eta (s)$ is the Dirichlet eta function, one finally has
$$\int_0^1 \frac{\ln^2 (1 + x)}{x} \, dx = -\frac{5}{4} \zeta (3) + \frac{3}{2} \zeta (3) = \frac{1}{4} \zeta (3),$$
as expected. 
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