While trying to evaluate a different integral, I met the following integral: $$I_n=\int\frac{(1-x^2)^{n-1}}{(1+x^2)^n}\mathrm{d}x$$ I was lost as to what to do with it, so I looked it up on wolfram alpha, and I got a result in terms of The Appell Hypergeometric Function $F_1$: $$I_n=xF_1\bigg(\frac12;1-n,n;\frac32;x^2,-x^2\bigg)+C$$ Where, for $|x|<1$, and $|y|<1$, $$F_1(a;b_1,b_2;c;x,y)=\sum_{m\geq0}\sum_{k\geq0}\frac{(a)_{m+k}(b_1)_m(b_2)_k}{m!k!(c)_{m+k}}x^my^k$$ With $$(x)_k=\frac{\Gamma(x+k)}{\Gamma(x)}$$ I found on Wikipedia that $$F_1(a;b1,b2;c;x,y)=\\\sum_{k\geq0}\frac{(a)_k(b_1)_k(b_2)_k(c-a)_k}{(c+k-1)_k(c)_{2k}}\frac{x^ky^k}{k!}\,_2F_1(a+k,b_1+k;c+2k;x)\,_2F_1(a+k,b_2+k;c+2k;y)$$ Which I thought might be useful.
I also know that $$\,_1F_0(1-n;;x^2)=(1-x^2)^{n-1}$$ Which gives $$I_n=\sum_{m\geq0}\frac{(1-n)_m}{m!}\int\frac{x^{2m}}{(1+x^2)^n}\mathrm{d}x$$ One could preform the substitution $u=1+x^2$: $$I_n=\frac12\sum_{m\geq0}\frac{(1-n)_m}{m!}\int\frac{(u-1)^{m-1/2}}{u^n}\mathrm{d}u$$ But this doesn't look any better.
Could anyone provide an alternative solution or give a proof of the result given by Wolfram?
Hint: With substitution $x=\tan t$ the integral is $$I_n=\int \cos^{n-1}2t\ dt$$ and it can be solved with integration by parts and recursive formula.