<s>[INST] Integral ${\large\int}_0^\infty\frac{\ln x}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}\ \mathrm dx$ Please help me to evaluate this integral:
$$
I={\large\int}_{0}^{\infty}{\ln\left(x\right) \over 1 + x}\,
\,\sqrt{\,x + \sqrt{\,1 + x^{2}\,}\, \over 1 + x^{2}\,}\,\,{\rm d}x.\tag1
$$
Mathematica could not evaluate it in a closed form. A numerical integration returned
$$I \approx 4.25314982536869548103063\ldots\,,\tag2$$
but neither
WolframAlpha nor ISC+ could find a plausible closed form for this.
 [/INST] \begin{align}
-\frac{1}{4}\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x
&=-\frac{1}{4}\int^\infty_0\frac{\ln{\sinh{x}}}{1+\sinh{x}}e^{x/2}{\rm d}x\\
&=-\frac{1}{2}\int^\infty_1\frac{\ln\left(\frac{x^2-x^{-2}}{2}\right)}{1+\frac{x^2-x^{-2}}{2}}{\rm d}x\\
&=\int^1_0\frac{\ln(1-x^4)-\ln(2x^2)}{x^4-2x^2-1}{\rm d}x\\
\end{align}
For simplicity's sake, let $\displaystyle\frac{1}{x^4-2x^2-1}=\sum^4_{k=1}\frac{c_k}{x-r_k}$. The integral becomes
$$I=-4\sum^4_{k=1}\int^1_0\frac{c_k\ln(1-x^4)-c_k\ln{2}-2c_k\ln{x}}{x-r_k}{\rm d}x$$
The second integral is
\begin{align}
-c_k\ln{2}\int^1_0\frac{1}{x-r_k}{\rm d}x=-c_k\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)
\end{align}
The third integral is
\begin{align}
-2c_k\int^1_0\frac{\ln{x}}{x-r_k}{\rm d}x
&=2c_k\int^{1/r_k}_0\frac{\ln(r_kx)}{1-x}{\rm d}x\\
&=-2c_k{\rm Li}_2\left(\frac{1}{r_k}\right)
\end{align}
Pluck these results back in.
$$I=4\sum^4_{k=1}c_k\left[\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)+2{\rm Li}_2\left(\frac{1}{r_k}\right)-\int^1_0\frac{\ln(1-x^4)}{x-r_k}{\rm d}x\right]$$
The remaining integral is
\begin{align}
&\ \ \ \ \ \int^1_0\frac{\ln(1-x^4)}{x-r_k}{\rm d}x\\
&=\sum_{j=1,-1,i,-i}\int^1_0\frac{\ln(1+jx)}{x-r_k}{\rm d}x\\
&=-\sum_{j=1,-1,i,-i}\int^{\frac{\lambda}{1-r_k}}_{-\frac{\lambda}{r_k}}\ln\left(1+\frac{j\lambda}{y}+jr_k\right)\frac{{\rm d}y}{y}\\
&=-\sum_{j=1,-1,i,-i}\int^{\frac{\lambda}{1-r_k}}_{-\frac{\lambda}{r_k}}\left[\ln\left(1+\frac{1+jr_k}{j\lambda}y\right)-\ln\left(\frac{y}{j\lambda}\right)\right]\frac{{\rm d}y}{y}\\
&=-\sum_{j=1,-1,i,-i}\int^{\frac{1+jr_k}{j-jr_k}}_{-\frac{1+jr_k}{jr_k}}\frac{\ln(1+y)}{y}-\frac{1}{y}\ln\left(\frac{y}{1+jr_k}\right){\rm d}y\\
&=-\sum_{j=1,-1,i,-i}\left[{\rm Li}_2\left(\frac{1+jr_k}{jr_k}\right)-{\rm Li}_2\left(\frac{1+jr_k}{jr_k-j}\right)+\frac{1}{2}\ln^2\left(-jr_k\right)-\frac{1}{2}\ln^2\left(j-jr_k\right)\right]
\end{align}
Final Result:
\begin{align}
\color\purple{\int^\infty_0\frac{\ln{x}}{1+x}\sqrt{\frac{x+\sqrt{1+x^2}}{1+x^2}}{\rm d}x
=4\sum^4_{k=1}c_k\left[\ln{2}\ln\left(\frac{1-r_k}{-r_k}\right)+2{\rm Li}_2\left(\frac{1}{r_k}\right)+\sum_{j=1,-1,i,-i}\left[{\rm Li}_2\left(\frac{1+jr_k}{jr_k}\right)-{\rm Li}_2\left(\frac{1+jr_k}{jr_k-j}\right)+\frac{1}{2}\ln^2\left(-jr_k\right)-\frac{1}{2}\ln^2\left(j-jr_k\right)\right]\right]}
\end{align}
where
\begin{align}
\frac{c_1}{x-r_1}&=\frac{1}{4\sqrt{2+2\sqrt{2}}\left(x-\sqrt{1+\sqrt{2}}\right)}\\
\frac{c_2}{x-r_2}&=-\frac{1}{4\sqrt{2+2\sqrt{2}}\left(x+\sqrt{1+\sqrt{2}}\right)}\\
\frac{c_3}{x-r_3}&=-\frac{1}{4\sqrt{2}\left(x+i\sqrt{\sqrt{2}-1}\right)}\\
\frac{c_4}{x-r_4}&=-\frac{1}{4\sqrt{2}\left(x-i\sqrt{\sqrt{2}-1}\right)}\\
\end{align}
</s>