How to Integrate $\int (\tan x)^{1/3}\,dx$

$$\tan x=u^3\implies x=\arctan(u^3)\implies dx=\frac{3u^2du}{1+u^6}$$

$$\implies\int\sqrt[3]{\tan x}dx=3\int\frac{u^3du}{1+u^6}=\frac32\int \frac{vdv}{1+v^3}(\text{ setting } u^2=v )$$

Now using Partial Fraction Decomposition, $$\frac{v}{1+v^3}=\frac A{1+v}+\frac {Bv+C}{1-v+v^2} $$

Multiply with $1+v^3$ and equate the constants & the coefficients of $v,v^2$

Now, $\displaystyle 1-v+v^2=\frac{4v^2-4v+4}4=\frac{(2v-1)^2+3}4$

Using Trigonometric substitution, set $2v-1=\sqrt3\tan\theta$

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