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While solving a problem I was stuck at this integral:

$$\int _{ 0 }^{ 1 }{ \frac { x{ \left( 1-x \right) }^{ \frac { 1 }{ 4 } } }{ { \left( 2-x \right) }^{ 2 } } dx } $$

Does this have a closed form? I tried to convert it into an integral that could be evaluated using beta function. But I couldn't. Please help me evaluate this.

Thanks!

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If we set $z=1-x$, then $z=u^4$, $$I=\int_{0}^{1}\frac{x(1-x)^{1/4}}{(2-x)^2}\,dx = \int_{0}^{1}\frac{z^{1/4}(1-z)}{(1+z)^2}\,dz = 4\int_{0}^{1}\frac{u^4(1-u^4)}{(1+u^4)^2}\,du$$ and the last integral can be computed through partial fraction decomposition, leading to: $$ I = \color{red}{\frac{1}{4} \left[-20+3\pi \sqrt{2}+6\sqrt{2} \log\left(1+\sqrt{2}\right)\right]}. $$

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  • $\begingroup$ Thanks for the solution! Could you elaborate the step of partial fractions? I was stuck over there itself. $\endgroup$ – Aditya Kumar Jul 4 '16 at 19:41
  • $\begingroup$ @AdityaKumar: $$(1+u^4)=(u^2+u\sqrt{2}+1)(u^2-u\sqrt{2}+1)$$ $\endgroup$ – Jack D'Aurizio Jul 4 '16 at 19:43
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    $\begingroup$ Ohk. I missed that. Thanks. $\endgroup$ – Aditya Kumar Jul 4 '16 at 19:44

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