Prove that $\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}$ We know that $$\int_0^1\frac{\ln (1+x)}{1+x^2}dx=\frac{\pi}8\ln 2,$$ but how about $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx?$$Prove that $$\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}.$$
 A: We will be using the following identities :
$$\int_0^1 x^{2n-1}\ln(1+x)dx=\frac{H_{2n}-H_n}{2n}\tag1$$
$$\sum_{n=1}^\infty x^n\frac{H_n}{n}=\frac12\ln^2(1-x)+\operatorname{Li}_2(x)\tag{2}$$
\begin{align}
I&=\int_0^1\frac{x\ln(1+x)}{1+x^2}\ dx=\sum_{n=1}^\infty(-1)^{n-1}\int_0^1x^{2n-1}\ln(1+x)\ dx\\
&\overset{(1)}{=}\sum_{n=1}^\infty(-1)^{n-1}\left(\frac{H_{2n}-H_n}{2n}\right)\\
&=\frac12\sum_{n=1}^\infty(-1)^n\frac{H_n}{n}-\sum_{n=1}^\infty(-1)^n\frac{H_{2n}}{2n}\\
&=\frac12\sum_{n=1}^\infty(-1)^n\frac{H_n}{n}-\Re\sum_{n=1}^\infty(i)^n\frac{H_n}{n}\\
&\overset{(2)}{=}\frac12\left(\frac12\ln^22+\operatorname{Li}_2(-1)\right)-\Re\left(\frac12\ln^2(1-i)+\operatorname{Li}_2(i)\right)\\
&=\frac12\left(\frac12\ln^22-\frac12\zeta(2)\right)-\left(\frac18\ln^22-\frac5{16}\zeta(2)\right)\\
&=\boxed{\frac18\ln^22+\frac1{16}\zeta(2)}
\end{align}

The proof of $(1)$ can be found here and the proof of $(2)$ follows from integrating the generating function $\sum_{n=1}^\infty H_n\ x^n=-\frac{\ln(1-x)}{1-x}$ .
A: An "elementary" solution.
\begin{align*}
J&=\int_0^1 \frac{x\ln(1+x)}{1+x^2}\,dx,K=\int_0^1 \frac{x\ln(1-x)}{1+x^2}\,dx\\
J+K&=\int_0^1 \frac{x\ln(1-x^2)}{1+x^2}\,dx\\
&\overset{y=\frac{1-x^2}{1+x^2}}=\frac{1}{2}\int_0^1 \frac{\ln\left(\frac{2y}{1+y}\right)}{1+y}\,dy\\
&=\frac{1}{2}\int_0^1 \frac{\ln 2}{1+y}\,dy+\frac{1}{2}\int_0^1 \frac{\ln y}{1+y}\,dy-\frac{1}{2}\int_0^1 \frac{\ln(1+y)}{1+y}\,dy\\
&=\frac{1}{4}\ln^2 2-\frac{1}{24}\pi^2\\
K-J&=\int_0^1 \frac{x\ln\left(\frac{1-x}{1+x}\right)}{1+x^2}\,dx\\
&\overset{y=\frac{1-x}{1+x}}=\int_0^1 \frac{\left(1-y\right)\ln y}{(1+y)(1+y^2)}\,dy\\
&=\int_0^1 \left(\frac{\ln y}{1+y}-\frac{y\ln y}{1+y^2}\right)\,dy\\
&=-\frac{\pi^2}{12}-\int_0^1 \frac{y\ln y}{1+y^2}\,dy\\
&\overset{z=y^2}=-\frac{\pi^2}{12}-\frac{1}{4}\int_0^1 \frac{\ln z}{1+z}\,dz\\
&=-\frac{1}{16}\pi^2
\end{align*}
Therefore,
\begin{align*}
J&=\frac{1}{2}\Big((J+K)-(K-J)\Big)\\
&=\frac{1}{2}\left(\frac{1}{4}\ln^2 2-\frac{1}{24}\pi^2+\frac{1}{16}\pi^2\right)\\
&=\boxed{\frac{1}{8}\ln^2 2+\frac{1}{96}\pi^2}
\end{align*}
NB:
I assume that,
\begin{align*}
\int_0^1 \frac{\ln x}{1+x}\,dx&=-\frac{\pi^2}{12}
\end{align*}
A: $\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,}
 \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace}
 \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack}
 \newcommand{\dd}{\mathrm{d}}
 \newcommand{\ds}[1]{\displaystyle{#1}}
 \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,}
 \newcommand{\ic}{\mathrm{i}}
 \newcommand{\mc}[1]{\mathcal{#1}}
 \newcommand{\mrm}[1]{\mathrm{#1}}
 \newcommand{\pars}[1]{\left(\,{#1}\,\right)}
 \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}}
 \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,}
 \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}}
 \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$
With $\ds{r \equiv 1 + \ic = 2^{1/2}\expo{\pi\ic/4}}$:
\begin{align}
&\int_{0}^{1}{x\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x =
\Re\int_{0}^{1}{\ln\pars{1 + x} \over x - \ic}\,\dd x =
\Re\int_{1}^{2}{\ln\pars{x} \over x - r}\,\dd x
\\[5mm] = &\
\Re\int_{1}^{1/2}{\ln\pars{1/x} \over 1/x - r}\,\pars{-\,{1 \over x^{2}}}\dd x =
-\,\Re\int_{1/2}^{1}{\ln\pars{x} \over x\pars{1 - rx}}\,\dd x
\\[5mm] = &\
-\int_{1/2}^{1}{\ln\pars{x} \over x}\,\dd x -
\Re\int_{1/2}^{1}{\ln\pars{x} \over 1 - rx}\,r\,\dd x
\\[5mm] = &\
{1 \over 2}\,\ln^{2}\pars{1 \over 2} -
\Re\int_{r/2}^{r}{\ln\pars{x/r} \over 1 - x}\,\dd x
\\[5mm] = &\
{1 \over 2}\,\ln^{2}\pars{1 \over 2} -
\Re\bracks{\ln\pars{1 - {r \over 2}}\ln\pars{1 \over 2} +
\int_{r/2}^{r}{\ln\pars{1 - x} \over x}\,\dd x}
\\[5mm] = &\
\Re\mrm{Li}_{2}\pars{r} - \Re\mrm{Li}_{2}\pars{r \over 2}
\\[1cm] = &\
{1 \over 2}\bracks{%
\mrm{Li}_{2}\pars{1 - \ic} + \mrm{Li}_{2}\pars{1 - {1 \over \ic}}}
\label{1}\tag{1}
\\[5mm] - &\
{1 \over 2}\braces{%
\mrm{Li}_{2}\pars{{1 \over 2} + {1 \over 2}\,\ic} +
\mrm{Li}_{2}\pars{1 - \bracks{{1 \over 2} + {1 \over 2}\,\ic}}}
\label{2}\tag{2}
\end{align}

\eqref{1} can be evaluated by means of
Landen Identity while \eqref{2} is evaluated with
Euler Reflection Formula.

Namely,
\begin{align}
&\int_{0}^{1}{x\ln\pars{1 + x} \over 1 + x^{2}}\,\dd x
\\[5mm] = &\
{1 \over 2}\bracks{%
-\,{1 \over 2}\,\ln^{2}\pars{i}} -
{1 \over 2}\braces{{\pi^{2} \over 6} -
\ln\pars{{1 \over 2} + {1 \over 2}\,\ic}
\ln\pars{1 -\bracks{{1 \over 2} + {1 \over 2}\,\ic}}}
\\[5mm] = &\
{\pi^{2} \over 16} - {\pi^{2} \over 12} +
{1 \over 2}\,\verts{\ln\pars{{1 \over 2} + {1 \over 2}\,\ic}}^{2} =
-\,{\pi^{2} \over 48} +
{1 \over 2}\,\verts{-\,{1 \over 2}\,\ln\pars{2} + {\pi \over 4}\,\ic}^{2}
\\[5mm] = &\
-\,{\pi^{2} \over 48} +
{1 \over 2}\,\bracks{{1 \over 4}\,\ln^{2}\pars{2} + {\pi^{2} \over 16}} =\
\bbox[#ffe,15px,border:1px dotted navy]{\ds{%
{\pi^{2} \over 96} + {\ln^{2}\pars{2} \over 8}}}
\end{align}
A: Put
\begin{equation*}
f(s) = \int_{0}^{1}\dfrac{x\ln(s+x)}{1+x^2}\, dx.
\end{equation*}
We want to determine $f(1)$. After differentiation we have
\begin{gather*}
f'(s) = \int_{0}^{1}\dfrac{x}{(s+x)(1+x^2)}\, dx = -\int_{0}^{1}\dfrac{s}{(s^2+1)(s+x)}\, dx +\int_{0}^{1}\dfrac{sx+1}{(s^2+1)(x^2+1)}\, dx  =
\\[2ex]
-\dfrac{s\ln(1+s)}{1+s^2}+\dfrac{\ln s}{1+s^2}+\dfrac{s\ln 2 }{2(1+s^2)}+\dfrac{{\pi}}{4(1+s^2)}.
\end{gather*}
Now we integrate wrt $s$ between $0$ and $1$. That yields
\begin{equation*}
f(1)-f(0) = -f(1)+f(0) +\dfrac{\ln^2(2)}{4}+\dfrac{{\pi}^2}{16}.
\end{equation*}
Consequently
\begin{equation*}
f(1)= f(0) +\dfrac{\ln^2(2)}{8} + \dfrac{{\pi}^2}{32}. \tag{1}
\end{equation*}
But
\begin{equation*}
f(0) = \int_{0}^{1}\dfrac{x\ln(x)}{1+x^2}\, dx = \sum_{k=0}^{\infty}\int_{0}^{1}(-1)^kx^{2k+1}\ln x\, dx =\sum_{k=0}^{\infty}\dfrac{(-1)^{k+1}}{4(k+1)^{2}} = -\dfrac{\pi^2}{48}.
\end{equation*}
Finally we substitute that into (1).
\begin{equation*}
f(1) = \dfrac{\ln^2(2)}{8} + \dfrac{{\pi}^2}{96}.
\end{equation*}
A: Let $$I=\int_0^1\frac{x\ln(1+x)}{1+x^2}dx$$
and $$I(a)=\int_0^1\frac{x\ln(1+ax)}{1+x^2}dx.$$
Note that $I(1)=I$ and $I(0)=0.$
$$I'(a)=\int_0^1\frac{x^2}{(1+x^2)(1+ax)}dx$$
$$=-\frac{\pi}{4}\frac{1}{1+a^2}+\frac12\ln(2)\frac{a}{1+a^2}+\frac{\ln(1+a)}{a}-\frac{a\ln(1+a)}{1+a^2}.$$
$$\Longrightarrow I=-\frac{\pi}{4}\int_0^1\frac{da}{1+a^2}+\frac12\ln(2)\int_0^1\frac{a}{1+a^2}da+\int_0^1\frac{\ln(1+a)}{a}da-\int_0^1\frac{a\ln(1+a)}{1+a^2}da$$
$$=-\frac{\pi}{4}\left(\frac{\pi}{4}\right)+\frac12\ln(2)\left(\frac12\ln(2)\right)+\frac{\pi^2}{12}-I$$
$$\Longrightarrow I=\frac{\pi^2}{96}+\frac18\ln^2(2).$$
