Evaluating $\int _0^1\frac{\ln ^2\left(x^2+1\right)}{x+1}\:dx$ using real methods I've stumbled upon that interesting integral here, the OP managed to transform the integral into something more approachable using contour integration and proved that
$$\int _0^1\frac{\ln ^2\left(x^2+1\right)}{x+1}\:dx\:=\:-\pi G+\frac{5}{2}\zeta \left(3\right)+\frac{2}{3}\ln ^3\left(2\right)-\frac{\pi ^2}{24}\ln \left(2\right)$$
I can't come up with a way to attack this one with only real methods, i'd appreciate any help possible.
 A: Integrate by parts
$$\int _0^1\frac{\ln ^2(x^2+1)}{x+1}~dx
=\ln^32 -4I \tag1$$
where $I$ is given below, along with its twin $J$
$$I=\int_0^1\frac{x\ln(1+x)\ln(x^2+1)}{1+x^2}~dx,\>\>\>\>\>
J=\int_0^1\frac{x\ln(1-x)\ln(x^2+1)}{1+x^2}~dx $$
Then, evaluate their sum and difference as follows
\begin{align}
J+I=& \int_0^1\frac{x\ln(1-x^2)\ln(x^2+1)}{1+x^2}~{dx}
\>\>\>\>\>\>\> \left( x^2\to\frac{1-x}{1+x} \right) \\
=& \frac12\int_0^1 \frac{\ln^2\frac{1+x}2}{1+x}dx
+\frac{\ln2}2 \int_0^1 \frac{\ln x}{1+x}dx+\frac14\int_0^1 \frac{\ln^2(1+x)}{x}dx\\
=&\frac16\ln^32-\frac{\pi^2}{24}\ln2+\frac1{16}\zeta(3)\\
\\
J-I=& \int_0^1\frac{x\ln\frac{1-x}{1+x}\ln(x^2+1)}{1+x^2}~{dx} \>\>\>\>\>\>\> \left( \frac{1-x}{1+x} \to x\right) \\
=& \>\ln2\int_0^1 \frac{\ln x}{1+x}dx
+\frac{\ln2}2  \int_0^1 \frac{\ln(1+x^2)}{x} \overset{x^2\to x}{dx}
+\frac14 \int_0^1 \frac{\ln^2(1+x^2)}{x} 
\overset{x^2\to x}{dx}\\
& -2\int_0^1 \frac{\ln x\ln(1+x)}{1+x}\overset{ibp}{dx} - \int_0^1 \frac{\ln (1+x)\ln(1+x^2)}{x}dx\\
= & \>\frac{3\ln2}4\int_0^1 \frac{\ln x}{1+x}dx
+\frac98 \int_0^1 \frac{\ln^2(1+x)}{x} dx 
- P 
=\frac9{32}\zeta(3) - \frac{\pi^2}{16}\ln2-P\\
\end{align}
where $P$ and its twin $Q$ are
$$P= \int_0^1 \frac{\ln (1+x)\ln(1+x^2)}{x}dx,\>\>\>\>\>
Q= \int_0^1 \frac{\ln (1-x)\ln(1+x^2)}{x}dx
$$
Similarly, their sum and difference are (see here for evaluation details)
\begin{align}
&Q+P=\int_0^1 \frac{\ln (1-x^2)\ln(1+x^2)}{x}dx=-\frac5{16}\zeta(3)\\
&Q-P=\int_0^1 \frac{\ln \frac{1-x}{1+x} \ln(1+x^2)}{x}dx=-\pi G- \frac7{4}\zeta(3)\\
\end{align}
which results in $P= \frac{\pi}{2}G-\frac{33}{32}\zeta(3) $. Substitute into $I-J$ and $I+J$ above to get
$$
I= \frac{\pi}4 G-\frac5{8}\zeta(3)
+ \frac{\pi^2}{96}\ln2+ \frac1{12}\ln^32
$$
and then into (1) to obtain
$$\int _0^1\frac{\ln ^2(1+x^2)}{1+x}\>dx=-\pi G+\frac{5}{2}\zeta (3)-\frac{\pi ^2}{24}\ln2 +\frac{2}{3}\ln^32 $$
