Generalized logarithm integral $\int^1_0 \frac{\log(1+x)}{1+a^2x^2}\, dx$ How to prove that 
$$\int^1_0 \frac{\log(1+x)}{1+a^2x^2}\, dx= \frac{1}{a} \Im \left( \chi_2(ia) - \operatorname{Li}_2\left( \frac{1+ai}{2}\right)\right)\,\,\, a> 0$$
where $\Im $ is the imaginary part , 
$\chi$ is the Legendre chi function , $\operatorname{Li}_2$ is the dilogarithm .
For the special case $a=1$ we have 
$$\int^1_0 \frac{\log(1+x)}{1+x^2}dx = \frac{\pi}{8}\log(2)$$
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$\ds{\int_{0}^{1}{\ln\pars{1 + x} \over 1 + a^{2}x^{2}}\,\dd x:\ {\large ?}.\qquad
     a \in {\mathbb R}}$.

I believe the proposed OP solution $\ds{\pars{~{1 \over a}\,\Im\pars{\chi_2\pars{\ic a}
      -{\rm Li}_{2}\pars{1\ +\ a\ic \over 2}}~}}$ is not correct. Albeit it agrees at
  $\ds{a = 1}$, we can evaluate it for several values and they yield different values between LHS and RHS.

\begin{align}&\color{#c00000}{\int_{0}^{1}%
{\ln\pars{1 + x} \over 1 + a^{2}x^{2}}\,\dd x}
=\Im\int_{0}^{1}{\ln\pars{1 + x} \over -\ic + \verts{a}x}\,\dd x
=\Im\int_{1}^{2}{\ln\pars{x} \over -\ic - \verts{a} + \verts{a}x}\,\dd x
\\[3mm]&=\Im\bracks{-\,{1 \over 1 + \ic\verts{a}}
\int_{1}^{2}{\ln\pars{x} \over 1 - \verts{a}x/\pars{\ic + \verts{a}}}\,\dd x}
\\[3mm]&=\Im\bracks{-\,{1 \over 1 + \ic\verts{a}}
\int_{1}^{2}{\ln\pars{x} \over 1 - \mu x}\,\dd x}\,,
\qquad\qquad
\mu \equiv {\verts{a} \over \ic + \verts{a}}
\end{align}

With
  $\ds{\mu x\equiv t\ \imp\ x = {t \over \mu}\quad\mbox{and}\quad
     \dd x={\ic + \verts{a} \over \verts{a}}\,\dd t}$
  \begin{align}&\color{#c00000}{\int_{0}^{1}%
{\ln\pars{1 + x} \over 1 + a^{2}x^{2}}\,\dd x}
={1 \over \verts{a}}\,
\Im\bracks{-\int_{\mu}^{2\mu}%
{\ln\pars{t/\mu} \over 1 - t}\,\dd t}
\\[3mm]&={1 \over \verts{a}}\,\Im\bracks{%
\ln\pars{1 - 2\mu}\ln\pars{2} - \int_{\mu}^{2\mu}{\ln\pars{1 - t} \over t}\,\dd t}
\\[3mm]&={1 \over \verts{a}}\,\Im\bracks{%
\ln\pars{1 - 2\mu}\ln\pars{2} + {\rm Li}_{2}\pars{2\mu} - {\rm Li}_{2}\pars{\mu}}
\end{align}

$$\color{#c00000}{\int_{0}^{1}%
{\ln\pars{1 + x} \over 1 + a^{2}x^{2}}\,\dd x}
={1 \over \verts{a}}\,\Im\bracks{%
\ln\pars{\ic - \verts{a} \over \ic + \verts{a}}\ln\pars{2}
+{\rm Li}_{2}\pars{2\verts{a} \over \ic + \verts{a}}
-{\rm Li}_{2}\pars{\verts{a} \over \ic + \verts{a}}}
$$

\begin{align}&\color{#66f}{\large\int_{0}^{1}%
{\ln\pars{1 + x} \over 1 + a^{2}x^{2}}\,\dd x}
\\[3mm]&=\color{#66f}{\large2\ln\pars{2}\,{\arctan\pars{a} \over a}
+{1 \over \verts{a}}\,\Im\bracks{%
{\rm Li}_{2}\pars{2\verts{a} \over \ic + \verts{a}}
-{\rm Li}_{2}\pars{\verts{a} \over \ic + \verts{a}}}}
\end{align}

A: A related problem. Note that, the Legendre chi function is related to the polylogarithm by the identity

$$ \chi_\nu(z) = \frac{1}{2}\left[\operatorname{Li}_\nu(z) - \operatorname{Li}_\nu(-z)\right]. $$  

