Solving the integral $\int_{0}^{1} \frac{s \ln(1+s) - t \ln(1+t)}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s$ For $t \in [0,1]$, let
$$f(t) = \int_{0}^{1} \frac{s \ln(1+s) - t \ln(1+t)}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s.$$
Is it possible to evaluate this integral to obtain an explicit expression for $f(t)$? This integral is very similar to the one solved in this answer, but unfortunately the technique used there of splitting up the integral does not seem to yield obviously tractable integrals (at least for Mathematica!).
Edit
This integral arose from a research problem in the theory of orthogonal polynomials, which involved finding the equilibrium measure for a particular weight function, $w(x) = \exp(-|x| \ln(1+|x|))$. The extra factor of 1 in the argument of the logarithm ensures smoothness of this weight function near zero.
Here is where I can get to using the method from the linked post.
\begin{align}
    f(t) &= \int_{0}^{1} \dfrac{s \ln(1+s) - t \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, \mathrm{d} s,\\
    &= \underbrace{\int_{0}^{1} \dfrac{s \ln(1+s) - s \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, \mathrm{d} s}_{\equiv I_{1}(t)} + \underbrace{\int_{0}^{1} \dfrac{s \ln(1+t) - t \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, \mathrm{d} s}_{\equiv I_{2}(t)}.
\end{align}
For $I_{2}(t)$, we have
\begin{align}
    I_{2}(t) &= \ln(1+t) \int_{0}^{1} \dfrac{s - t}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, \mathrm{d} s,\\
    &= \ln(1+t) \int_{0}^{1} \dfrac{1}{\sqrt{1-s^{2}}(s + t)} \, \mathrm{d} s,\\
    &= \dfrac{2 \ln(1+t)}{\sqrt{1-t^{2}}} \mathrm{arctanh}{\sqrt{\dfrac{1-t}{1+t}}}.
\end{align}
For $I_{1}(t)$, change variables to $x = \sqrt{1-s^{2}}$, and define $b \equiv \sqrt{1-t^{2}}$. Then we have
\begin{equation}
    I_{1} = \int_{0}^{1} \dfrac{\ln(1+\sqrt{1-x^{2}}) - \ln(1+\sqrt{1-b^{2}})}{b^{2} - x^{2}} \, \mathrm{d} x.
\end{equation}
But whereas in the linked post this integral was solvable by Mathematica, in this case apparently we have no such luck.
 A: Continue with
\begin{align}
    & \int_{0}^{1} \dfrac{s \ln(1+s) - t \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s\\
   = &\ \underset{=J}{\int_{0}^{1} \dfrac{s \ln(1+s) - s \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s} + {\int_{0}^{1} \dfrac{s \ln(1+t) - t \ln(1+t)}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s}\\
   =&\ J + \frac{2 \ln(1+t)}{\sqrt{1-t^{2}}} \mathrm{arctanh}{\sqrt{\dfrac{1-t}{1+t}}}
\end{align}
and evaluate $J$ below with the variable changes
$ x^2=\frac{1-s}{1+s}$ and $ y^2=\frac{1-t}{1+t}$
\begin{align}
J &= \int_{0}^{1} \frac{s\ln\frac{1+s}{1+t}}{\sqrt{1-s^{2}}(s^{2} - t^{2})} \, {d} s
= \int_{0}^{1} \frac{\ln\frac{1+s}{1+t}}{2\sqrt{1-s^{2}} }\left( \frac 1{s-t} + \frac1{s+t}  \right){d} s\\
&=\frac{1+y^2}2\int_0^1 \frac{\ln\frac{1+y^2}{1+x^2}}{y^2-x^2}
 + \frac{\ln\frac{1+y^2}{1+x^2}}{1-y^2x^2}\ dx\\
\end{align}
where
\begin{align}
\int_0^1 \frac{\ln\frac{1+y^2}{1+x^2}}{y^2-x^2}dx
&= \frac1y \Re\bigg( \text{Li}_2\left(\frac{y+1}{y-i}\right)-\text{Li}_2\left(\frac{y-1}{y+i}\right) \bigg) \\
\int_0^1 \frac{\ln\frac{1+y^2}{1+x^2}}{1-y^2x^2}dx
&= \frac1y \Re\bigg( \text{Li}_2\left(\frac{1+y}{1-iy}\right)-\text{Li}_2\left(\frac{1-y}{1+iy}\right)\bigg)- \frac1y \ln y \ \ln\frac{1-y}{1+y}
\end{align}
