Solving the integral $\int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s$ For $t \in [0,1]$, let
$$f(t) = \int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}\,\mathrm{d}s.$$
Is it possible to evaluate this integral to obtain an explicit expression for $f(t)$? Mathematica and Maple both fail to give an answer.

Split the integral into two
\begin{align}
I=& \int_{0}^{1} \frac{s \ln{s} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds=I_1+I_2
\end{align}
where, with $b = \sqrt{1-t^2}$
\begin{align}
I_1= & \int_{0}^{1} \frac{s \ln{s} - s \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds
\overset{x=\sqrt{1-s^2}}=\frac12 \int_{0}^{1}
 \frac{\ln\frac{1-x^2}{1-b^2}}{b^2-x^2}dx\\
=&\ \frac1{4b}\bigg[\text{Li}_2\left(\frac{b-1}{b+1}\right)-\text{Li}_2\left(\frac{b+1}{b-1}  \right)\bigg]\\
\\
I_2= & \int_{0}^{1} \dfrac{s \ln{t} - t \ln{t}}{(s^{2} - t^{2})\sqrt{1-s^{2}}}ds\\
=& \int_{0}^{1} \dfrac{\ln{t}}{(s +t)\sqrt{1-s^{2}}}ds=\frac{2\ln t}{\sqrt{1-t^2}}\tanh^{-1}\sqrt{\frac{1-t}{1+t}}\\
\end{align}
Thus
\begin{align}
I=& \frac1{4\sqrt{1-t^2}}\left[ \text{Li}_2\bigg(\frac{\sqrt{1-t^2}-1}{\sqrt{1-t^2} +1}\bigg)-\text{Li}_2\bigg(\frac{\sqrt{1-t^2} +1}{\sqrt{1-t^2}-1}  \bigg)
+ 8\ln t\tanh^{-1}\sqrt{\frac{1-t}{1+t}}\right]
\end{align}
