Need help with $\int_0^1 \frac{\ln(1+x^2)}{1+x} dx$ Can the definite integral
$$\int_0^1\dfrac{\ln(1+x^2)}{1+x}dx$$
be evaluated using the technique of “differentiation under integral sign”. I don't want a complete solution, just the parameter would do.
PS: An alternative approach (preferably simple) would also be welcome as long as it doesn't involve contour integration.
 A: $\newcommand{\angles}[1]{\left\langle\, #1 \,\right\rangle}
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\begin{align}
&\color{#66f}{\large\int_{0}^{1}{\ln\pars{1 + x^{2}} \over 1+x}\,\dd x}
=\ln^{2}\pars{2} - \int_{0}^{1}\ln\pars{1 + x}\,{2x \over 1 + x^{2}}\,\dd x
\\[5mm]&=\ln^{2}\pars{2}
-2\,\color{#c00000}{\Im\int_{0}^{1}{\ln\pars{1 + x} \over 1 - \ic x}\,\dd x}
=\color{#66f}{\large-\,{\pi^{2} \over 48} + {3 \over 4}\,\ln^{2}\pars{2}}
\approx {\tt 0.1547}
\end{align}

With $\ds{\mu \equiv {\ic \over 1 + \ic}=\half\,\pars{1 + \ic}}$:

\begin{align}
&\color{#c00000}{\Im\int_{0}^{1}{\ln\pars{1 + x} \over 1 - \ic x}\,\dd x}
=\Im\int_{1}^{2}{\ln\pars{x} \over 1 + \ic - \ic x}\,\dd x
=\Im\bracks{-\ic\int_{1}^{2}{\ln\pars{x} \over 1 - \ic x/\pars{1 + \ic}}
\,{\ic\,\dd x \over 1 + \ic}}
\\[5mm]&=\Im\bracks{-\ic\int_{1}^{2}{\ln\pars{x} \over 1 - \mu x}\,\mu\,\dd x}
=\Im\bracks{-\ic\int_{\mu}^{2\mu}{\ln\pars{x/\mu} \over 1 - x}\,\dd x}
\\[5mm]&=\Im\bracks{\left.\ic\ln\pars{1 - x}\ln\pars{x \over \mu}
\right\vert_{x\ =\ \mu}^{x\ =\ 2\mu}
-\ic\int_{\mu}^{2\mu}\ln\pars{1 - x}\,{1/\mu \over x/\mu}\,\dd x}
\\[5mm]&=\Im\braces{\ic\ln\pars{1 - 2\mu}\ln\pars{2} + \ic\int_{\mu}^{2\mu}\
\underbrace{\bracks{-\,{\ln\pars{1 - x} \over x}}}
_{\ds{=\ \color{#c00000}{{\rm Li}_{2}'\pars{x}}}}\ \,\dd x}
\\[5mm]&=\Re\,{\rm Li}_{2}\pars{1 + \ic} - \Re\,{\rm Li}_{2}\pars{1 + \ic \over 2}
\end{align}
A: (Edit: Just noticed that the OP specifically requested a hint instead of a complete solution. My apologies.)
Start by integrating by parts:
$$\begin{align}
\mathcal{I}
&:=\int_{0}^{1}\frac{\ln{\left(1+x^2\right)}}{1+x}\,\mathrm{d}x\\
&=\left[\ln{(1+x)}\ln{(1+x^2)}\right]_{0}^{1}-\int_{0}^{1}\frac{2x\ln{(1+x)}}{1+x^2}\,\mathrm{d}x\\
&=\ln^2{(2)}-2\int_{0}^{1}\frac{x\ln{(1+x)}}{1+x^2}\,\mathrm{d}x\\
&=\ln^2{(2)}-2\mathcal{J}{(1)},\\
\end{align}$$
where we've introduced the parameter $a$ by defining the function $\mathcal{J}{(a)}:=\int_{0}^{1}\frac{x\ln{(1+ax)}}{1+x^2}\,\mathrm{d}x$.
A: 
I don't want a complete solution, just the parameter would do.

Try
\begin{equation}
I(a)=\int_0^1\frac{\ln\left(1+a^2x^2\right)}{1+x}\, dx
\end{equation}

No upvote yet? Okay...

Differentiating w.r.t. $a$, we have
\begin{align}
I'(a)&=\int_0^1\left[\frac{2ax^2}{(1+x)\left(1+a^2x^2\right)}\right]\, dx\\
&=\frac{2a}{1+a^2}\int_0^1\frac{1}{1+x}\,dx-\frac{2a}{1+a^2}\int_0^1\frac{1-x}{1+a^2x^2}\,dx\\
&=\frac{2a\ln2}{1+a^2}-\frac{2\arctan a}{1+a^2}+\frac{\ln\left(1+a^2\right)}{a\left(1+a^2\right)}\\
\end{align}
then integrating back
\begin{align}
I(a)&=\ln\left(1+a^2\right)\ln2-\arctan^2(a)+\int\left[\frac{\ln\left(1+a^2\right)}{a}-\underbrace{\frac{a\ln\left(1+a^2\right)}{1+a^2}}_{({\Large\color{red}{\star}})}\right]\,da\\
&=\ln\left(1+a^2\right)\ln2-\arctan^2(a)-\frac{1}{2}{\rm{Li}}\left(-a^2\right)-\frac{1}{4}\ln^2\left(1+a^2\right)+C\\
\end{align}
For $({\Large\color{red}{\star}})$ we use integration by parts once by taking $u=\ln\left(1+a^2\right)$.
Now, I'm sure you can you take it from here.
