jacks wrote:$(a)\;$ Evaluation of $\displaystyle \int_{0}^{\sqrt{2}-1}\frac{\ln(1+x^2)}{1+x}dx$
For the first integral and somewhat quick. The solution will actually involve poylogarithms and basic complex things.
\begin{align*}
\int_{0}^{\sqrt{2}-1} \frac{\ln \left ( 1+x^2 \right )}{1+x} \, {\rm d}x&= \int_{0}^{\sqrt{2}-1} \frac{\ln \left ( 1-i^2 x^2 \right )}{1+x} \, {\rm d}x\\
&= \int_{0}^{\sqrt{2}-1} \frac{\ln (1-ix) + \ln (1+ix)}{1+x} \, {\rm d}x\\
&= \int_{0}^{\sqrt{2}-1} \frac{\ln (1-ix)}{1+x} \, {\rm d}x + \int_{0}^{\sqrt{2}-1} \frac{\ln (1+ix)}{1+x} \, {\rm d}x\\
&=-{\rm Li}_2 \left ( \frac{1-i}{2} \right ) +{\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) + \frac{i}{4} \pi \ln \left [ \left ( 1+i \right ) - i \sqrt{2} \right ] \\
& \quad \; -{\rm Li}_2 \left ( \frac{1+i}{2} \right ) + {\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) - \frac{i}{4} \pi \ln \left [ (1-i) + i \sqrt{2} \right ] \\
&= - \left [ -i \mathcal{G} + \frac{\pi^2}{48} - \frac{1}{2}\left ( -\frac{\ln 2}{2} + \frac{i \pi}{4} \right )^2 \right ] + {\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) \\
&\quad \; -\left [ i \mathcal{G} + \frac{\pi^2}{48} - \frac{1}{2} \left ( -\frac{\ln 2}{2} - \frac{i \pi}{4} \right )^2 \right ] +{\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) + \frac{\pi^2}{16} \\
&= \frac{\ln^2 2}{4} - \frac{5 \pi^2}{48} + \frac{\pi^2}{16} + {\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) +{\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) \\
&= \frac{\ln^2 2}{4} - \frac{\pi^2}{24} + {\rm Li}_2 \left ( 1 - \frac{1-i}{\sqrt{2}} \right ) +{\rm Li}_2 \left ( 1 - \frac{1+i}{\sqrt{2}} \right ) \approx 0.0173049
\end{align*}
Side note: One interesting identity regarding the dilog function is the following:
$$\Re{ \left[{\rm Li}_{2}\left(\frac{1}{2}+iq\right) \right]}=\frac{{\pi}^{2}}{12}-\frac{1}{8}{\ln^2{\left(\frac{1+4q^2}{4}\right)}}-\frac{{\arctan^2{(2q)}}}{2}$$
where $q \in \mathbb{Q}$ and there is a similar one for the imaginary part which I will post later.
Addendum: In general it holds that:
\begin{align*}
\int_{0}^{1} \frac{\log (1+ax)}{1+x} \, {\rm d}x &\overset{u=1+x}{=\! =\! =\!} \int_{1}^{2} \frac{\log \left ( 1-a+au \right )}{u} \, {\rm d}u \\
&=\int_{1}^{2} \frac{\log \left [ \left ( 1-a \right ) \left ( 1+ \frac{a}{1+a} u \right ) \right ]}{u} \, {\rm d}u \\
&= \log \left ( 1-a \right ) \int_{1}^{2} \frac{{\rm d}u}{u} +\int_{1}^{2} \frac{\log \left ( 1+\frac{a}{1-a} u \right )}{u} \, {\rm d}u \\
&= \log 2 \log (1-a) - \sum_{n=1}^{\infty} \frac{1}{n} \left ( -\frac{a}{1-a} \right )^n\int_{1}^{2} u^{n-1} \, {\rm d}u\\
&= \log 2 \log (1-a) - \sum_{n=1}^{\infty} \frac{1}{n^2} \left ( -\frac{a}{1-a} \right )^n \left ( 2^n-1 \right ) \\
&= \log 2 \log (1-a) -{\rm Li}_2 \left ( -\frac{2a}{1-a} \right ) + {\rm Li}_2 \left ( -\frac{a}{1-a} \right )
\end{align*}
Hence we conclude that:
$$\bbox[blue, 2pt]{{\color{white}{\int_{0}^{1}\frac{\log (1+ax)}{1+x} \, {\rm d}x = \log 2 \log (1-a) -{\rm Li}_2 \left ( -\frac{2a}{1-a} \right ) + {\rm Li}_2 \left ( - \frac{a}{1-a} \right )}}}$$
for $a \neq 1$, while for $a=1$ we get that $\displaystyle \int_0^1 \frac{\log (1+x)}{1+x}\, {\rm d}x = \frac{\log^2 2}{2}$.