Showing that $\int_{0}^{1}{\sqrt{1-x^4}\over 1+x^4}dx={\pi\over 4}$

Let $y = x^2$ and $z = \sqrt{y^{-1} - y}$, we have $$\begin{align} \int_0^1 \frac{\sqrt{1-x^4}}{1+x^4}dx = & \int_0^1 \frac{x\sqrt{x^{-2} - x^2}}{x^2(x^{-2}+x^2)}dx = \int_0^1 \frac{\sqrt{x^{-2} - x^2}}{(x^{-2}+x^2)}\frac{dx}{x}\\ = & \frac12 \int_0^1 \frac{\sqrt{y^{-1}-y}}{y^{-1}+y}\frac{dy}{y} = -\frac12 \int_{y=0}^1 \frac{\sqrt{y^{-1}-y}}{y^{-1}+y}\frac{d(y^{-1} - y)}{y^{-1}+y}\\ = & \int_0^\infty \frac{z^2 dz}{z^4+4} = \int_0^\infty \frac{z^2 dz}{(z^2+2)^2 - (2z)^2} = \int_0^\infty \frac{z^2 dz}{(z^2-2z+2)(z^2+2z+2)}\\ = & \frac14 \int_0^\infty\left(\frac{z}{z^2-2z+2} - \frac{z}{z^2+2z+2}\right)dz\\ = & \frac14 \int_0^\infty \left[\frac12\log\left(\frac{z^2-2z+2}{z^2+2z+2}\right)' + \left(\frac{1}{z^2-2z+2} + \frac{1}{z^2+2z+2}\right)\right] dz \end{align} $$ The first piece contributes $\displaystyle\;\frac18 \left[\log\left(\frac{z^2-2z+2}{z^2+2z+2}\right)\right]_0^\infty\;$ which clearly vanishes.

For the second piece, substitute $z$ by $-z$ in its second term, we get:

$$\int_0^1 \frac{\sqrt{1-x^4}}{1+x^4}dx = \frac14 \int_{-\infty}^\infty\frac{dz}{z^2-2z+2} = \frac14 \int_{-\infty}^\infty\frac{dz}{(z-1)^2+1} = \frac{\pi}{4}$$

You were so close! $$\begin{align}\int_0^{\frac{\pi}4}\sqrt{\cot2u}\,du&=\frac12\int_0^{\frac{\pi}2}\sqrt{\cot v}\,dv\tag{1}\\ &=\frac14\cdot2\int_0^{\frac{\pi}2}\cos^{1/2}v\sin^{-1/2}v\,dv\tag{2}\\ &=\frac14\text{B}\left(\frac34,\frac14\right)\tag{3}\\ &=\frac14\frac{\Gamma\left(\frac34\right)\Gamma\left(\frac14\right)}{\Gamma(1)}\tag{4}\\ &=\frac14\frac{\pi}{\sin\frac{\pi}4}\tag{5}\\ &=\frac{\pi}{2\sqrt2}\tag{6}\end{align}$$ EDIT: Some annotations:
$(1)$ Let $2u=v$
$(2)$ $\cot\theta=\frac{\cos\theta}{\sin\theta}$
$(3)$ Recognizing the trigonometric form of the Beta function
$(4)$ Using the relationship between the Beta and Gamma functions
$(5)$ The reflection formula for the Gamma function. We might equally have used the duplication formula $$\Gamma\left(\frac14\right)\Gamma\left(\frac34\right)=\sqrt{\pi}2^{1-2\left(\frac14\right)}\Gamma\left(2\left(\frac14\right)\right)=\sqrt{\pi}\sqrt2\sqrt{\pi}=\pi\sqrt2$$ $(6)$ Of course $\sin\frac{\pi}4=\frac1{\sqrt2}$. At this point we're home free and we can substitute the value of the integral into where you were stuck and arrive at $$\int_0^1\frac{\sqrt{1-x^4}}{1+x^4}dx=\frac{\sqrt2}2\frac{\pi}{2\sqrt2}=\frac{\pi}4$$