Answer
$2\tan \sqrt{x}+C$
Work Step by Step
Let $u=\sqrt{x}$. Then $u=x^\frac{1}{2}$, so $du=\frac{1}{2}x^{-\frac{1}{2}}dx=\frac{1}{2\sqrt{x}}dx$. This means that $\frac{1}{\sqrt{x}}dx=2du$.
$\int\frac{\sec^2\sqrt{x}}{\sqrt{x}}dx$
$=\int\sec^2\sqrt{x}*\frac{1}{\sqrt{x}}dx$
$=\int\sec^2u*2 du$
$=2\tan u+C$
$=2\tan \sqrt{x}+C$
