A slightly problematic integral $\int{1/(x^4+1)^{1/4}} \, \mathrm{d}x$

Hint:

This can be written as : $$\int \frac{x^4dx}{x^5\left(1+\frac{1}{x^4}\right)^{1/4}}$$ Now substitute $1+\frac{1}{x^4}=t^4$ $$\implies t^3dt=-\frac{1}{x^5}dx$$ and $$x^4=\frac{1}{t^4-1}$$ to get $$\int \frac{t^2dt}{1-t^4}$$ Now use partial fractions.

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 $$I = \frac{1}{2}\ln \left|\frac{t-1}{t+1}\right|-\frac{1}{2}\tan^{-1}(t)+\mathcal{C}$$