TITLE: How can I find  $\int\frac1{\sqrt[4]{1+x^4}}\mathrm dx$?
QUESTION [7 upvotes]: My question is, how can I evaluate the following integral?
$$\int\frac1{\sqrt[4]{1+x^4}}\mathrm dx$$
Thanks.

REPLY [7 votes]: Integral
$$\begin{align*}
\int\frac{1}{(1+x^4)^{1/4}}dx&=\int \frac{1}{x(1+1/x^4)^{1/4}}dx\\
&=\int \frac{x^4}{x^5(1+1/x^4)^{1/4}}dx.
\end{align*}$$
Substitution: $z^4=(1+1/x^4)$
$4z^3 dz=-4\frac{1}{x^5}dx$
Therefore,
$$\begin{align*}
\int\frac{1}{(1+x^4)^{1/4}}dx&=\int \frac{x^4}{x^5(1+1/x^4)^{1/4}}dx\\
&=-\int\frac{z^2}{(z^4-1)}dz\\
&=-\frac{1}{2}(\frac{1}{z^2-1}+\frac{1}{z^2+1})dz\\
&=-\frac{1}{2}\ln\left|\frac{1-z}{1+z}\right|+\arctan z+ C
\end{align*}$$
where $z=(1+1/x^4)^{1/4}$.