Prove that $\int_0^1\frac{x\ln (1+x)}{1+x^2}dx=\frac{\pi^2}{96}+\frac{\ln^2 2}{8}$

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*}

This is a solution using complex analysis.

The integral \begin{equation*} \int_{0}^{1}\dfrac{\ln x}{1+x}\, dx = \sum_{k=0}^{\infty}(-1)^k\int_{0}^{1}x^k\ln x\, dx = \sum_{k=1}^{\infty}(-1)^{k}\dfrac{1}{k^2} = -\dfrac{{\pi}^2}{12} \tag{1} \end{equation*} will be useful.

Put \begin{equation*} I = \int_{0}^{1}\dfrac{x\ln(1+x)}{1+x^2}\, dx. \end{equation*} After integration by parts we have \begin{equation*} I = \dfrac{1}{2}\ln^22 - \dfrac{1}{2}I_1 \tag{2} \end{equation*} where \begin{equation*} I_1 = \int_{0}^{1}\dfrac{\ln(1+x^2)}{1+x}\, dx. \end{equation*} To evaluate $I_1$ we integrate $\dfrac{\log(1+z^2)}{1+z}$ over $\gamma_{1}+\gamma_{2}$, where $\gamma_{1}$ is the imaginary axis from $0$ to $i$ and $\gamma_{2}$ the unit circle from $i$ to $1$ clockwise. With $\log z$ we mean $\ln|z|+i\arg z$, where $-\pi <\arg z < \pi$. (The singularity in $i$ can be handled by a flanking movement along a ''quater circle'' with a shrinking radius.)

Put \begin{equation*} I_{2} = \int_{\gamma_{1}}\dfrac{\log(1+z^2)}{1+z}\, dz = \int_{0}^{1}\dfrac{\ln(1-t^2)}{1+it}i\, dt = \int_{0}^{1}\dfrac{\ln(1-t^2)}{1+t^2}i(1-it)\, dt. \end{equation*} Since $I_1$ is real we are only interested in the real part of $I_2$. After the substitution $t = \dfrac{1}{s+1}$ we get \begin{gather*} {\rm{Re}} I_2 = \int_{0}^{1}\dfrac{t\ln(1-t^2)}{1+t^2}\, dt = I + \int_{0}^{1}\dfrac{t\ln(1-t)}{1+t^2}\, dt = I + \int_{0}^{\infty}\dfrac{\ln s-\ln(s+1)}{(s+1)(s^2+2s+2)}\, ds = \\[2ex] I + \int_{0}^{\infty}\dfrac{\ln s}{(s+1)(s^2+2s+2)}\, ds -\int_{0}^{\infty}\dfrac{\ln(s+1)}{(s+1)((s+1)^2+1)}\, ds . \end{gather*} The first integral can be evaluated by integrating $\dfrac{\log^2z}{(z+1)(z^2+2z+2)}$ along a keyhole contour. We get the value $\dfrac{1}{8}\ln^2 2 -\dfrac{\pi^2}{32}$ via residue calculus.

In the the second integral we substitute $s+1$ to $\dfrac{1}{\sqrt{u}}$. If we combine that with (1) we get \begin{equation*} -\dfrac{1}{4}\int_{0}^{1}\dfrac{\ln u}{1+u}\, du = \dfrac{\pi^2}{48}. \end{equation*} Consequently \begin{equation*} {\rm{Re}} I_2 = I +\dfrac{1}{8}\ln^2 2 -\dfrac{\pi^2}{32} -\dfrac{\pi^2}{48} = I +\dfrac{1}{8}\ln^2 2 - \dfrac{5\pi^2}{96}. \tag{3} \end{equation*} Now to the unit circle. \begin{gather*} I_{3} = \int_{\gamma_{2}}\dfrac{\log(1+z^2)}{1+z}\, dz = -\int_{0}^{\pi/2}\dfrac{\log(1+e^{i2t})}{1+e^{it}}ie^{it}\, dt =\\[2ex] -\int_{0}^{\pi/2}\dfrac{\ln(2\cos t) +it}{2\cos(t/2)}i(\cos(t/2)+i\sin(t/2))\,dt. \end{gather*} We extract the real part and finally use the substitution $s = \cos t$ and (1). \begin{gather*} {\rm{Re}} I_3 = \dfrac{1}{2}\int_{0}^{\pi/2}\left(t + \tan(t/2)\ln(2\cos t)\right)\, dt=\\[2ex] \dfrac{\pi^2}{16} + \dfrac{1}{2}\int_{0}^{1}\dfrac{\ln 2}{1+s}\, ds + \dfrac{1}{2}\int_{0}^{1}\dfrac{\ln s}{1+s}\, ds = \dfrac{\pi^2}{16} +\dfrac{1}{2}\ln^22 - \dfrac{\pi^2}{24} = \dfrac{1}{2}\ln^22 + \dfrac{\pi^2}{48}.\tag{4} \end{gather*} We substitute (3) and (4) in (2). \begin{equation*} I = \dfrac{1}{2}\ln^22 - \dfrac{1}{2}\left(I +\dfrac{1}{8}\ln^2 2 - \dfrac{5\pi^2}{96}+ \dfrac{1}{2}\ln^22 + \dfrac{\pi^2}{48}\right) = \dfrac{3}{16}\ln^22 +\dfrac{\pi^2}{64} -\dfrac{1}{2}I. \end{equation*} Finally \begin{equation*} I = \dfrac{1}{8}\ln^22 + \dfrac{\pi^2}{96}. \end{equation*}

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 $\sum_{n=1}^\infty H_n\ x^n=-\frac{\ln(1-x)}{1-x}$ .

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*}

$\newcommand{\bbx}[1]{\,\bbox[8px,border:1px groove navy]{\displaystyle{#1}}\,} \newcommand{\braces}[1]{\left\lbrace\,{#1}\,\right\rbrace} \newcommand{\bracks}[1]{\left\lbrack\,{#1}\,\right\rbrack} \newcommand{\dd}{\mathrm{d}} \newcommand{\ds}[1]{\displaystyle{#1}} \newcommand{\expo}[1]{\,\mathrm{e}^{#1}\,} \newcommand{\ic}{\mathrm{i}} \newcommand{\mc}[1]{\mathcal{#1}} \newcommand{\mrm}[1]{\mathrm{#1}} \newcommand{\pars}[1]{\left(\,{#1}\,\right)} \newcommand{\partiald}[3][]{\frac{\partial^{#1} #2}{\partial #3^{#1}}} \newcommand{\root}[2][]{\,\sqrt[#1]{\,{#2}\,}\,} \newcommand{\totald}[3][]{\frac{\mathrm{d}^{#1} #2}{\mathrm{d} #3^{#1}}} \newcommand{\verts}[1]{\left\vert\,{#1}\,\right\vert}$ 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}