I came upon the problem to get a formula depending on $k\ge 1$ for the definite integral $$\int_0^1 \frac{t^{2k}}{(1-t^2)^k} dt$$
Please note that 1 is a singularity of the integrand.
I tried on WolframAlpha online to get a formula for the integral without the bounds zero and one which to my surprise could be expressed by the hypergeometric function as above. The evaluation of the hypergeometric function fails for t=1.
$$\int \frac{t^{2k}}{(1-t^2)^k} dt=\frac{t^{2k+1}}{2k+1}{_2}F{_1}(k,k+1/2;k+3/2;t^2)+C$$
As I am not current any more in the integration method of complex analysis when there is a singularity. I need a hint if that can be applied in case of this singularity.
The only other method I know is integration by parts. Maybe that would also result in the above equation.
Or kind of a recurrence equation in k ?
If $k\geq 1$ the singularity in the left neighbourhood of $1$ is non-integrable, if $k\leq -\frac{1}{2}$ the singularity in the right neighbourhood of zero is non-integrable. On the other hand, assuming $k\in\left(-\frac{1}{2},1\right)$ we have
$$ \int_{0}^{1}\frac{t^{2k}}{(1-t^2)^k}\,dt = \frac{1}{2}\int_{0}^{1} u^{k-1/2}(1-u)^{-k}\,du = \frac{1}{2}\,B\left(1-k,k+\frac{1}{2}\right) $$ which equals $$ \frac{\Gamma(1-k)\Gamma\left(k+\tfrac{1}{2}\right)}{2\Gamma\left(\tfrac{3}{2}\right)} =\frac{\sqrt{\pi}\,\Gamma\left(k+\tfrac{1}{2}\right)}{\Gamma(k)\sin(\pi k)}=\frac{2\pi\,\Gamma(2k)}{4^k\,\Gamma(k)^2\sin(\pi k)}.$$