Solve the integral using beta functions : $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx$ I was solving questions on beta and gamma functions and then I came across this question.
$$ \int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,dx $$
Generally in questions of beta functions integrals of the type $\int x^n(1−x^m) \, dx$ can be solved by substituting $x^m = t$, but in this case substitution is unconventional because if I put $1+x^4=t$, then I will not get the required form of beta function
I tried substituting $x^2 = \tan t$ and also tried substituting $x^2 = \cos t$ but on simplifying each of them further I could not reduce them to standard form of beta function.
 A: We can use the Binomial Theorem in the form
$$
(1+x)^{-\alpha}=\sum_{k=0}^\infty\frac{(-1)^k\Gamma(k+\alpha)}{\Gamma(\alpha)\Gamma(k+1)}x^k\tag1
$$
to get
$$
\begin{align}
\int_0^1\frac{(1-x^4)^{3/4}}{(1+x^4)^2}\,\mathrm{d}x
&=\frac14\int_0^1\frac{(1-x)^{3/4}}{(1+x)^2}x^{-3/4}\,\mathrm{d}x\tag2\\
&=\frac14\int_0^1(1-x)^{3/4}x^{-3/4}\sum_{k=0}^\infty(k+1)(-x)^k\,\mathrm{d}x\tag3\\
&=\frac14\sum_{k=0}^\infty(-1)^k(k+1)\int_0^1(1-x)^{3/4}x^{k-3/4}\,\mathrm{d}x\tag4\\
&=\frac14\sum_{k=0}^\infty(-1)^k(k+1)\frac{\Gamma\!\left(\frac74\right)\Gamma\left(k+\frac14\right)}{\Gamma(k+2)}\tag5\\
&=\underbrace{\frac14\Gamma\!\left(\frac74\right)\Gamma\!\left(\frac14\right)\vphantom{\sum_{k=0}^\infty}}_{\frac{3\sqrt2}{16}\pi}\underbrace{\sum_{k=0}^\infty\frac{(-1)^k\Gamma\left(k+\frac14\right)}{\Gamma\left(\frac14\right)\Gamma(k+1)}}_{(1+1)^{-1/4}}\tag6\\
&=\frac{3\sqrt[4]2}{16}\pi\tag7
\end{align}
$$
Explanation:
$(2)$: substitute $x\mapsto x^{1/4}$
$(3)$: apply $(1)$ with $\alpha=2$ and use $(k+1)\Gamma(k+1)=\Gamma(k+2)$
$(4)$: collect the integral for the Beta Function
$(5)$: evaluate the integral
$(6)$: rearrange factors and use $(k+1)\Gamma(k+1)=\Gamma(k+2)$
$(7)$: $\frac14\Gamma\!\left(\frac74\right)\Gamma\!\left(\frac14\right)
=\frac14\cdot\frac34\Gamma\!\left(\frac34\right)\Gamma\!\left(\frac14\right)
=\frac3{16}\pi\csc\left(\frac\pi4\right)
=\frac{3\sqrt2}{16}\pi$
$\phantom{(7)\text{:}}$ apply $(1)$ with $\alpha=\frac14$ and $x=1$ 
A: Apply the substitution $x^4 = \frac{1-u}{1+u}$. Then
$$ I := \int_{0}^{1} \frac{(1-x^4)^{3/4}}{(1+x^4)^2} \, dx
= \frac{1}{2^{9/4}} \int_{0}^{1} u^{3/4}(1-u)^{-3/4} \, du
= \frac{1}{2^{9/4}} B(\tfrac{7}{4},\tfrac{1}{4})
= \frac{3\pi}{2^{15/4}}.$$

Explanations. First substitute $x^4 = t$ to obtain $ I = \frac{1}{4} \int_{0}^{1} \frac{(1-t)^{3/4}t^{-3/4}}{(1+t)^2} \, dx$. To transform this integral into the form of beta integral, we notice that 


*

*$-1$, $0$ and $1$ are the only branch points/poles of the integrand,

*$0$, $1$, $\infty$ are the only branch points/poles of the integrand of $\int_{0}^{1} u^{\alpha-1}(1-u)^{\beta-1} \, du$.


This leads to consider a Möbius transformation $t = f(u) = \frac{au+b}{cu+d}$ satisfying either 


*

*$f(\infty) = -1$, $f(0) = 0$ and $f(1) = 1$ and $ad-bc > 0$, or

*$f(\infty) = -1$, $f(0) = 1$ and $f(1) = 0$ and $ad-bc < 0$. 


The first condition yields  $f(u) = \frac{u}{2-u}$ and the second condition yields $f(u) = \frac{1-u}{1+u}$. Both can be used, since one condition is simply transformed to the other by $z \mapsto 1-z$.
