<s>[INST] Evaluate $\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$ Any idea on how to solve the following definite integral?

$$\int_0^1\frac{\ln{(x^2+1)}}{x+1}dx$$

I have tried to parameterize the integral like $\ln{(a^2x^2+1)}$ or $\ln{(x^2+a^2)}$, which don't seem to work.
 [/INST] Not a full answer (yet)
I'll try to provide a solution which doesn't use any special functions, only the well known "classic" series.
$$I=\int_0^1 \frac{\ln{(x^2+1)}}{x+1}dx=\sum_{k,n=0}^\infty \frac{(-1)^{k+n}}{k+1} \int_0^1 x^{2(k+1)+n}dx=\sum_{k,n=0}^\infty \frac{(-1)^{k+n}}{(k+1)(2k+n+3)}$$
Let's use partial fractions:
$$\frac{1}{(k+1)(2k+n+3)}=\frac{a}{k+1}+\frac{b}{2k+n+3} \\ 2a+b=0 \\ (n+3)a+b=1 \\ (n+1)a=1 \\ a=\frac{1}{n+1} \\ b=-\frac{2}{n+1}$$
So we have:
$$I=\sum_{k,n=0}^\infty \frac{(-1)^n}{n+1} \frac{(-1)^k}{k+1}-2 \sum_{k,n=0}^\infty \frac{(-1)^n}{n+1} \frac{(-1)^k}{2k+n+3}=\ln^2 2-2\sum_{n=0}^\infty \frac{(-1)^n}{n+1} \sum_{k=0}^\infty \frac{(-1)^k}{2k+n+3}$$
The second series can be represented the following way, separating even and odd terms:
$$\sum_{n=0}^\infty \frac{(-1)^n}{n+1} \sum_{k=0}^\infty \frac{(-1)^k}{2k+n+3}=\sum_{l=0}^\infty \frac{1}{2l+1} \sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+3}-\sum_{l=0}^\infty \frac{1}{2l+2} \sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+4}$$
Now let's deal with the inner series separately:
1) $$\sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+3}=(-1)^{l+1} \sum_{k=l+1}^\infty \frac{(-1)^k}{2k+1}=(-1)^{l+1} \left( \sum_{k=0}^\infty \frac{(-1)^k}{2k+1}- \sum_{k=0}^l \frac{(-1)^k}{2k+1} \right)= \\ =(-1)^{l+1} \left( \frac{\pi}{4}- \sum_{k=0}^l \frac{(-1)^k}{2k+1} \right)$$
The first term is the well known Leibniz series for $\pi$.
2) $$\sum_{k=0}^\infty \frac{(-1)^k}{2(k+l)+4}=\frac{(-1)^{l+1}}{2} \sum_{k=l+1}^\infty \frac{(-1)^k}{k+1}=\frac{(-1)^{l+1}}{2} \left( \sum_{k=0}^\infty \frac{(-1)^k}{k+1}- \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)= \\ =\frac{(-1)^{l+1}}{2} \left(\ln 2- \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)$$
The first terms of each can now be summed w.r.t. $l$:
$$\sum_{l=0}^\infty \frac{1}{2l+1} (-1)^{l+1} \frac{\pi}{4}=-\frac{\pi^2}{16}$$
$$\frac{1}{2} \sum_{l=0}^\infty \frac{1}{l+1} \frac{(-1)^{l+1}}{2} \ln 2=-\frac{\ln^2 2}{4}$$
At this point we have:
$$I=\frac{\ln^2 2}{2}+\frac{\pi^2}{8}-\sum_{l=0}^\infty (-1)^l \left(\frac{2}{2l+1} \sum_{k=0}^l \frac{(-1)^k}{2k+1}-\frac{1}{2}\frac{1}{l+1} \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)$$

Well, I tried. Now it seems that there's at least two similar questions with some good answers, but I'll still leave this attempt here in case it could be finished.

We need to prove that:
$$\sum_{l=0}^\infty (-1)^l \left(\frac{2}{2l+1} \sum_{k=0}^l \frac{(-1)^k}{2k+1}-\frac{1}{2}\frac{1}{l+1} \sum_{k=0}^{l} \frac{(-1)^k}{k+1} \right)=\frac{7 \pi^2}{48}-\frac{\ln^2 2}{4}$$
which seems like a long shot, even though numerically it's true.
Mostly I just wanted to show how the squared logarithm and $\pi$ appear in the answer.
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