<s>[INST] Evaluate $\int\limits_0^\infty \frac{\ln^2(1+x)}{1+x^2}\ dx$ This problem was already solved here (in different closed form).

But how can you prove $\ \displaystyle\int\limits_0^\infty\frac{\ln^2(1+x)}{1+x^2}\ dx=2\Im\left(\operatorname{Li}_3(1+i)\right)\ $

Where $\displaystyle \operatorname{Li}_3(x)=\sum_{n=1}^\infty\frac{x^n}{n^3}\ $ is the the trilogarithm.
 [/INST] The following solution is by Cornel Valean:
\begin{gather*}
 \int_0^\infty\frac{\ln^2(1+x)}{1+x^2}\textrm{d}x\overset{x= \frac1y}{=}\int_0^\infty\frac{\ln^2\left(\frac{y}{1+y}\right)}{1+y^2}\textrm{d}y\\
 \overset{\frac{y}{1+y}=x}{=}\int_0^1\frac{\ln^2(x)}{x^2+(1-x)^2}\textrm{d}x\\
\left\{\text{write $\frac{1}{x^2+(1-x)^2}=\Im \frac{1+i}{1-(1+i)x}$}\right\}\\
 =\Im \int_0^1\frac{(1+i)\ln^2(x)}{1-(1+i)x}\textrm{d}x\\
 =2\ \Im\{\operatorname{Li}_{3}(1+i)\}. 
 \end{gather*}
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