Integral $\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2}$, may involve beta function Evaluate the integral: $$\int_0^1 \frac{(1-x^4)^{3/4}}{(1+x^4)^2} dx$$
I have used substitutions like $x^4 = u$, or $x^{-4} = u$. After many hours, I came up with $\frac{2}{1+x^4} = u$ which reduces the integral to :
$$\int_1^2 \frac{2^{3/4}}{8} \left(\frac{u-1}{2-u}\right)^{3/4} du$$
and then taking $\frac{u-1}{2-u} = z$ gives:
$$\frac{2^{3/4}}{8} \int_0^\infty \frac{z^{3/4} }{(1+z)^2} dz = \color{red}{ \frac{2^{3/4}}{8}\beta(\tfrac{7}{4},\tfrac{1}{4})}$$
Last answer comes from using:
$$\int_0^\infty \frac{x^{m-1}}{(1+x)^{m+n}} dx = \beta(m,n)$$
This answer seems to be correct as online integral calculator gives $0.7005...$ as answer.
What can be alternate approaches
 A: Alternative approaches? The integral is a hypergeometric functions, and I would immediately try to get it because then I could always use some identities to simplify if needed.
Let's use the substitution $x^4=v$, which the OP rejected.
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{1}{4} \int_0^1 v^{-3/4} (1-v)^{3/4}(1+v)^{-2} dv$$
We know the general integral for the hypergeometric function:
$$\mathrm {B} (b,c-b)\,_{2}F_{1}(a,b;c;z)=\int _{0}^{1}x^{b-1}(1-x)^{c-b-1}(1-zx)^{-a}\,dx\qquad \Re (c)>\Re (b)>0$$
In this case $a=2$, $b=1/4$, $c=2$, so we get:
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{1}{4} \mathrm {B} \left(\frac14 , \frac74 \right) {_2 F_1} \left(2,\frac14; 2;-1 \right)$$
Using some simplifications:
$$\frac{1}{4}\mathrm {B} \left(\frac14 , \frac74 \right)=\frac{1}{4}\Gamma \left( \frac14 \right) \Gamma \left(\frac74 \right)=\Gamma \left(\frac54 \right)\Gamma \left(\frac74 \right)$$
$${_2 F_1} \left(2,\frac14; 2;-1 \right)={_1 F_0} \left(\frac14; ;-1 \right)=(1-(-1))^{-1/4}=\frac{1}{2^{1/4}}$$
(It's known that ${_1 F_0} \left(a; ;x \right)=(1-x)^{-a}$).
So we get:
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{1}{2^{1/4}}\Gamma \left(\frac54 \right)\Gamma \left(\frac74 \right)$$
Which is the same as the OP's result, but obtained in a more simple way, using the known properties of hypergeometric functions. 
This can be further simplified by using the reflection formula for the Gamma function (thanks to @Szeto for reminding me):
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{3}{16 \cdot 2^{1/4}}\Gamma \left(\frac14 \right)\Gamma \left(\frac34 \right)$$
$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{3\pi}{16 \cdot 2^{1/4} \sin \frac{\pi}{4}}$$


$$\int_0^1 (1-x^4)^{3/4}(1+x^4)^{-2} dx=\frac{3\pi \cdot 2^{1/4}}{16}$$

A: Easiest way to solve this is by taking 4 substitutions
$$x^4=u \\
\dfrac 1u=z \\
z-1=t\\
\dfrac t2=v$$
After these the integral will come out in the form of the last identity op has mentioned
I know my formatting and illustration is weak,so kindly someone pls form a answer on basis of this and post it as it may help people searching for the same Q
