<s>[INST] Integral $\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c}$ Hi I am trying to prove this result $$
I:=\int_0^1 \log x \frac{(1+x^2)x^{c-2}}{1-x^{2c}}dx=-\left(\frac{\pi}{2c}\right)^2\sec ^2 \frac{\pi}{2c},\quad c>1.
$$
Thanks.  Since $x\in[0,1]
$ we can write$$
I=\sum_{n=0}^\infty \int_0^1 \log x (1+x^2) x^{c-2+2cn}dx.
$$
since $\sum_{n=0}^\infty x^n= (1-x)^{-1} , |x| < 1.$  Simplifying
$$
I=\sum_n \int_0^1 \log x\, x^{c-2+2n} dx+\sum_n \int_0^1 \log x \, x^{c+2cn}\, dx.
$$
Now we can write
$$
I=-\frac{1}{4}\psi_1 \left(\frac{c+1}{2}\right)-\frac{1}{4} \psi_1\left( \frac{3+c}{2}\right)
$$
where I summed the results of the integrals using general result of $\int_0^1 \log x \, x^n \, dx=-\frac{1}{(n+1)^2}$.  Howeever this is not the result $\sec^2...$.  Thanks
The function $\psi$ is the polygamma function which is defined in general by
$$
\psi_m(z)=\frac{d^{m+1}}{dz^{m+1}} \log \Gamma(z)
$$
and $\Gamma(z)=(z-1)!$, for this case $m=1$.
 [/INST] Using the identity
\begin{align}
\psi(1-x) - \psi(x) = \pi \cot(\pi x)
\end{align}
then the derivative with respect to $x$ yields
\begin{align}
\psi_{1}(1-x) + \psi_{1}(x) = \pi^{2} \csc^{2}(\pi x).
\end{align}
Now the integral
\begin{align}
I &= \int_{0}^{1} \frac{ x^{c-2} (1+x^{2}) \ln(x) }{ 1 - x^{2c} } \, dx
\end{align}
is evaluate as follows.
\begin{align}
I &= \sum_{n=0}^{\infty} \, \int_{0}^{1} (1+x^{2}) \, x^{2cn+c-2} \ln(x) \, dx \\
&= - \sum_{n=0}^{\infty} \left[ \frac{1}{(2cn+c-1)^{2}} + \frac{1}{(2cn+c+1)^{2}} \right] \\
&= - \frac{1}{4 c^{2}} \left[ \psi_{1}\left(\frac{c-1}{2c}\right) + \psi_{1}\left( \frac{c+1}{2c}\right) \right] \\
&= - \left(\frac{\pi}{2c}\right)^{2} \csc^{2}\left(\frac{\pi}{2} + \frac{\pi}{2c}\right) \\
&= - \left(\frac{\pi}{2c}\right)^{2} \sec^{2}\left(\frac{\pi}{2c}\right).  
\end{align}
It can now be stated that
\begin{align}
\int_{0}^{1} \frac{ x^{c-2} (1+x^{2}) \ln(x) }{ 1 - x^{2c} } \, dx = - \left(\frac{\pi}{2c}\right)^{2} \sec^{2}\left(\frac{\pi}{2c}\right).
\end{align} 
</s>