∫ (tan^4√x sec^2√xdx)/(√x) equals

\(I=\int \frac{\tan^{4}\sqrt{x} \sec^{2}\sqrt{x}}{\sqrt{x}}dx\)

let $y=\tan\sqrt{x}$

$2dy=\frac{\sec^2\sqrt{x}}{\sqrt{x}}dx$

so $I=2\int y^4dy=\frac{2}{5}y^5+C$

$=\frac{2}{5}\tan^{5}\sqrt{x}+C$