TITLE: How to evaluate $\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx$
QUESTION [29 upvotes]: How to evaluate
  $$\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx$$

My attempt
$$\begin{align}\int_{0}^{1}{\frac{x\ln(1+x)\ln(1+x^2)}{1+x^2}}dx&=\frac{1}{4}\int_{0}^{1}{\ln(1+x)}d(\ln^2(1+x^2))\\ &=\frac{1}{4}(\ln(1+x)\ln^2(1+x^2)|_0^1-\int_{0}^{1}{\ln^2(1+x^2)}d\ln(1+x))\\ &=\frac{1}{4}(\ln^32-\int_{0}^{1}\frac{\ln^2(1+x^2)}{1+x}dx)\\
\end{align}$$
$$\begin {align} 
&\int_{0}^{1} \frac{\ln ^{2}\left(1+x^{2}\right)}{1+x} d x\\=&\int_{0}^{1} \frac{\ln ^{2}\left(1-y^{2}\right)}{1+i y} i d y-\int_{0}^{\frac{\pi}{2}} \frac{\ln ^{2}\left(1+e^{i 2 \theta}\right)}{1+e^{i \theta}} i e^{i \theta} d\theta\\
=&\int_{0}^{1} \frac{y \ln ^{2}\left(1-y^{2}\right)}{1+y^{2}} dy+\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \tan \left(\frac{\theta}{2}\right) \ln ^{2}(2 \cos \theta)d\theta+\int_{0}^{\frac{\pi}{2}} \theta \ln (2 \cos \theta)d\theta\\&-\frac{1}{2} \int_{0}^{\frac{\pi}{2}} \theta^{2} \tan \left(\frac{\theta}{2}\right)d\theta
 \end {align}$$

But how to evaluate
  $$\int_{0}^{1}\frac{\ln^2(1+x^2)}{1+x}dx$$

The answer from Mathematica is $$\frac{1}{96}(24\pi\mathbf{G}+\pi^2\ln2+8\ln^32-60\zeta(3))$$
$\mathbf{G}$ is Catalan's constant.
My answer

REPLY [3 votes]: Here is a solution similar to FDP's, by trying to build a relation with a similar integral we have:
$$I=\int _0^1\frac{x\ln \left(1-x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx,\:J=\int _0^1\frac{x\ln \left(1+x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx$$
$$I+J=\int _0^1\frac{x\ln \left(1-x^2\right)\ln \left(1+x^2\right)}{1+x^2}\:dx$$
$$=\frac{1}{2}\int _0^1\frac{\ln \left(1-x\right)\ln \left(1+x\right)}{1+x}\:dx$$
$$I+J=\frac{1}{16}\zeta \left(3\right)-\frac{1}{4}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{6}\ln ^3\left(2\right)$$
That last integral can be computed easily if one considers the algebraic identity $ab=\frac{1}{4}\left(a+b\right)^2-\frac{1}{4}\left(a-b\right)^2$.

$$I-J=\int _0^1\frac{x\ln \left(\frac{1-x}{1+x}\right)\ln \left(1+x^2\right)}{1+x^2}\:dx$$
$$=\int _0^1\frac{\ln \left(x\right)\ln \left(1+x^2\right)}{1+x}\:dx+\ln \left(2\right)\int _0^1\frac{\ln \left(x\right)}{1+x}\:dx-2\underbrace{\int _0^1\frac{\ln \left(x\right)\ln \left(1+x\right)}{1+x}\:dx}_{N}$$
$$-\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx-\ln \left(2\right)\int _0^1\frac{x\ln \left(x\right)}{1+x^2}\:dx+2\underbrace{\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx}_{K}$$
$$=\int _0^1\frac{\ln \left(x\right)\ln \left(1+x^2\right)}{1+x}\:dx-\frac{3}{8}\ln \left(2\right)\zeta \left(2\right)-\frac{9}{4}\int _0^1\frac{\ln \left(x\right)\ln \left(1+x\right)}{1+x}\:dx$$
$$+2\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx$$
$$=\int _0^1\frac{\ln \left(x\right)\ln \left(1+x^2\right)}{1+x}\:dx-\frac{3}{16}\zeta \left(3\right)$$
$$=-2\int _0^1\frac{x\ln \left(x\right)\ln \left(1+x\right)}{1+x^2}\:dx-\int _0^1\frac{\ln \left(1+x\right)\ln \left(1+x^2\right)}{x}dx-\frac{3}{16}\zeta \left(3\right)$$
$$=\frac{9}{32}\zeta \left(3\right)-\frac{3}{8}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_{2k}}{k^2}-\frac{1}{2}\sum _{k=1}^{\infty }\frac{\left(-1\right)^kH_k}{k^2}$$
$$=\frac{9}{32}\zeta \left(3\right)-\frac{3}{8}\ln \left(2\right)\zeta \left(2\right)+\frac{23}{32}\zeta \left(3\right)-\frac{\pi }{2}G+\frac{5}{16}\zeta \left(3\right)$$
$$I-J=\frac{21}{16}\zeta \left(3\right)-\frac{3}{8}\ln \left(2\right)\zeta \left(2\right)-\frac{\pi }{2}G$$
The integral $N$ can be computed quite simply by considering the substitution $x=\frac{1}{1+x}$.
As for the integral $K$ can be found calculated in a similar fashion in the book (Almost) Impossible Integrals, Sums, and Series page $\#98$.

We have:
\begin{alignat*}{4}
I & {}+{}  & J & {}={}  \frac{1}{16}\zeta \left(3\right)-\frac{1}{4}\ln \left(2\right)\zeta \left(2\right)+\frac{1}{6}\ln ^3\left(2\right) \\
 I & {}-{}  &  J & {}={}   \frac{21}{16}\zeta \left(3\right)-\frac{3}{8}\ln \left(2\right)\zeta \left(2\right)-\frac{\pi }{2}G \\
\end{alignat*}

So we conclude that:
$$I=\int _0^1\frac{x\ln \left(1-x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx=\frac{11}{16}\zeta \left(3\right)-\frac{5}{16}\ln \left(2\right)\zeta \left(2\right)-\frac{\pi }{4}G+\frac{1}{12}\ln ^3\left(2\right)$$
$$J=\int _0^1\frac{x\ln \left(1+x\right)\ln \left(1+x^2\right)}{1+x^2}\:dx=-\frac{5}{8}\zeta \left(3\right)+\frac{1}{16}\ln \left(2\right)\zeta \left(2\right)+\frac{\pi }{4}G+\frac{1}{12}\ln ^3\left(2\right)$$