$$ \int \frac{dx}{(1+x^4)^{1/4}}=\frac{1}{2}\bigg[\frac{1}{2}\log\Big|\frac{1+z}{1-z}\Big|-\tan^{-1}z\bigg]+C $$ Find $z$
Method 1
Set $x^2=\tan t\implies 2x.dx=\sec^2t.dt\implies dx=\dfrac{\sec^2t.dt}{2\sqrt{\tan t}}$
$$
\int\frac{dx}{(1+x^4)^{1/4}}=\int\frac{\sec^2t.dt}{2\sqrt{\tan t.\sec t}}=\int\frac{1}{2\cos t\sqrt{\sin t}}dt=\int\frac{\cos t}{2.\cos^2 t\sqrt{\sin t}}dt
$$
Set $y=\sin t\implies dy=\cos t.dt$
$$
I=\int\frac{dy}{2(1-y^2)\sqrt{y}}
$$
Set $z=\sqrt{y}\implies dz=\dfrac{dy}{2\sqrt{y}}$
$$
I=\int\frac{dz}{1-z^4}=\frac{1}{2}\int\frac{dz}{1-y^2}-\frac{1}{2}\int\frac{dz}{1+z^2}=\frac{1}{2}\bigg[\frac{1}{2}\log\Big|\frac{1+z}{1-z}\Big|-\tan^{-1}z\bigg]\\
\sec^2t=1+x^4\implies\cos^2t=\frac{1}{1+x^4}\implies\sin^2t=1-\frac{1}{1+x^4}=\frac{x^4}{1+x^4}\\
\implies\boxed{z=\sqrt{\sin t}=\frac{x}{(1+x^4)^{1/4}}}
$$
Method 2
$$
I=\int\frac{dx}{(1+x^4)^{1/4}}=\int\frac{dx}{x(1+\dfrac{1}{x^4})^{1/4}}=\int\frac{x^4.dx}{x^5(1+\dfrac{1}{x^4})^{1/4}}
$$
Set $1+\dfrac{1}{x^4}=z^4\implies 4z^3.dz=\dfrac{-4}{x^5}dx$
$$
I=\int\frac{-z^3.dz}{z(z^4-1)}=\int\frac{-z^2}{z^4-1}dz=\frac{-1}{2}\int\frac{z^2+1+z^2-1}{(z^2-1)(z^2+1)}dz\\
=\frac{-1}{2}\bigg[\int\frac{dz}{z^2-1}+\int\frac{dz}{z^2+1}\bigg]\\
=\frac{-1}{2}.\frac{1}{2}\log|\frac{z-1}{z+1}|-\frac{1}{2}\tan^{-1}z+C\\
=\frac{1}{2}\bigg[\frac{1}{2}\log|\frac{1+z}{1-z}-\tan^{-1}z|\bigg]+C'\\
\implies\boxed{z=\frac{(1+x^4)^{1/4}}{x}}
$$
This type of integrals are attempted to solve at Integral of type $\displaystyle \int\frac{1}{\sqrt[4]{x^4+1}}dx$, and Thanx @lab bhattacharjee.
Why do I seem to get different expression for $z$ in methods 1 and 2 ?
Note: My reference gives the solution $z=\dfrac{(1+x^4)^{1/4}}{x}$
Both solutions only differ by a constant and a mistake in sign because
$$\log\left|\frac{1+\frac{1}{z}}{1-\frac{1}{z}}\right| = \log\left|\frac{1+z}{1-z}\right|$$
and
$$\tan^{-1}\left(\frac{1}{z}\right) = \frac{\pi}{2} - \tan^{-1}(z)$$
which is consistent because there is a sign error in a method, I just can't immediately see where.