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

You are on the right track, but the remaining computations are tedious.  First, note that we have the factorizations $$t^4 + 1 = (t^2 - \sqrt{2} t + 1)(t^2 + \sqrt{2} t + 1),$$ and $$t^8 + 1 = (t^2 - \alpha t + 1)(t^2 + \alpha t + 1)(t^2 - \beta t + 1)(t^2 + \beta t + 1),$$ where $\alpha = \sqrt{2+\sqrt{2}}$ and $\beta = \sqrt{2 - \sqrt{2}} = \sqrt{2}/\alpha$.  These then admit partial fraction decomposition of the two integrands you obtained, from which we can then eventually arrive at an antiderivative.
Incidentally, the separation into two terms as you have done is not particularly useful.
An alternate approach would be to perform the partial fraction decomposition of $\dfrac{t^4}{t^8+1}$ into linear factors over $\mathbb C$, for which the antiderivative is easier to compute as these will be logarithms of the form $$\log (t - \zeta_{16}^{2k+1})$$ for $\zeta_{16} = e^{\pi i/8}$ a primitive $16^{\rm th}$ root of unity and $k = 0, 1, \ldots, 7$.  Then we recombine these appropriately to obtain an expression without complex numbers.
