A slightly problematic integral $\int{1/(x^4+1)^{1/4}} \, \mathrm{d}x$ Question. To find the integral of- $$\int {\frac{1}{(x^4+1)^\frac{1}{4}} \, \mathrm{d}x}$$
I have tried substituting $x^4+1$ as $t$, and as $t^4$, but it gives me an even more complex integral. Any help?

Let $$I = \int\frac{1}{(x^4+1)^{\frac{1}{4}}}dx$$
Put $x^2=\tan \theta,$ Then $2xdx = \sec^2 \theta d\theta$
So $$I = \int\frac{\sec^2 \theta}{\sqrt{\sec \theta}}\cdot \frac{1}{2\sqrt{\tan \theta}}d\theta = \frac{1}{2}\int\frac{1}{\cos \theta \sqrt{\sin \theta}}d\theta = \frac{1}{2}\int\frac{\cos \theta}{(1-\sin^2 \theta)\sqrt{\sin \theta}}d\theta$$
Now Put $\sin \theta = t^2\;,$ Then $\cos \theta d\theta = 2tdt$
So $$I = \int\frac{1}{1-t^4}dt = -\int\frac{1}{(t^2-1)(t^2+1)}dt = -\frac{1}{2}\int\left[\frac{1}{1-t^2}+\frac{1}{1+t^2}\right]dt$$
So $$I = \frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|-\frac{1}{2}\tan^{-1}(t)+\mathcal{C}$$
