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How to evaluate this integral :

$$I = \int^1_0 \frac{x^{p-1}}{(1-x)^p(1+qx)^p}dx $$

$$q>0 , -1/2<p<1$$ My attemp : $$$$ Let $x=\cos^2(y)-\frac{1}{q}\sin^2(y)$ $$dx = -2\left({1+\frac{1}{q}}\right)\sin(y)\cos(y) dy $$

therfore:

$$I = \frac{2q^{p-1}}{(1+q)^{2p-1}}\int^{\arctan \sqrt{q}}_{0}\;\;\;\frac{\left[{\cos^2(y)-\frac{1}{q}\sin^2(y)}\right]^{p-1}}{\sin^{2p-1}(y)\cos^{2p-1}(y)}dy$$

this proplem in book : Table of Integrals, Series, and Products Eighth Edition page 320

In book : $$I = \int^1_0 \frac{x^{p-1}}{(1-x)^p(1+qx)^p}dx = \frac{2\Gamma\left({p+\frac{1}{2}}\right)\Gamma(1-p)\cos^{2p}\left({\arctan\left({\sqrt q}\right)}\right)\sin\left({(2p-1)\arctan{\sqrt{q}}}\right)}{\sqrt{\pi}(2p-1)\sin\left({\arctan{\sqrt{q}}}\right)}$$

Is there a way to evaluate this integration?

Delta
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    In the seventh edition the numerator of the integrand is $x^{p-1/2}$. With this modification and using Euler's integral representation for the hypergeometric function ${}2F{1}(a,b;c;z)$, the value of the integral is $$\frac{2 \Gamma(p+1/2) \Gamma(1-p)}{\sqrt{\pi}} , {}2F{1}(p,p+1/2;3/2;-q). $$ The result then follows from the second part of this answer. – Random Variable Sep 23 '24 at 21:00
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    For $q>0$ and $0\leq p <1$ Mathematica gives $\pi \csc (\pi p) , _2F_1(p,p;1;-q)$. Mathematica also states that the integral doesn't converge if $-1/2<p<0$. – JimB Sep 23 '24 at 21:01

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