Answer
$$\frac{5}{4} \sec \left(x^{4 / 5}\right)+c$$
Work Step by Step
Given $$\int x^{-1 / 5} \sec \left(x^{4 / 5}\right) \tan \left(x^{4 / 5}\right) dx$$
Let
$$ u=x^{4 / 5} \ \ \ \ \Rightarrow \ \ \ du =\frac{5}{4} x^{-1/5}dx$$
Then
\begin{aligned} \int x^{-1 / 5} \sec \left(x^{4 / 5}\right) \tan \left(x^{4 / 5}\right) dx&=\frac{5}{4} \int \sec (u) \tan (u) d u \\ &=\frac{5}{4} \sec u +c\\
&=\frac{5}{4} \sec \left(x^{4 / 5}\right)+c\end{aligned}
