<s>[INST] Evaluation of $\int\sqrt[4]{\tan x}dx$ Evaluation of $\displaystyle \int\sqrt[4]{\tan x}dx$
$\bf{My\; Try}::$ Let $\tan x = t^4\;\;,$ Then $\sec^2 xdx = 4t^3dt$. So $\displaystyle dx = \frac{4t^3}{1+t^8}dt$
So Integral Convert into $\displaystyle 4\int\frac{t^4}{1+t^8}dt = 2\int \frac{(t^4+1)+(t^4-1)}{t^8+1}dt$
So Integral is $\displaystyle 2\int\frac{t^4+1}{t^8+1}dt+2\int\frac{t^4-1}{t^4+1}dt$
Now How can I solve after that
Help me
Thanks
 [/INST] In general, with $t=\sqrt[n]{\tan x}$
\begin{align}
&\int\sqrt[n]{\tan x}\>dx=n\int \frac{t^n}{1+t^{2n}}dt\\
=&\>\frac14 
 \sum_{k=1}^{2n} (-1)^{k-1}\left(\sin a_k \ln(x^2-2x\cos a_k+1)-2\cos a_k \tan^{-1}\frac{\sin a_k}{x-\cos a_k}\right) +C
\end{align}
where $a_k=\frac{(2k-1)\pi}{2n}$.
</s>