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

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!

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: $$I = \color{red}{\frac{1}{4} \left[-20+3\pi \sqrt{2}+6\sqrt{2} \log\left(1+\sqrt{2}\right)\right]}.$$
• @AdityaKumar: $$(1+u^4)=(u^2+u\sqrt{2}+1)(u^2-u\sqrt{2}+1)$$ Jul 4 '16 at 19:43