Integrate the following function with respect to $x:$
$\frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}}$

(N/A) Let $I = \int \frac{\tan ^{4} \sqrt{x} \sec ^{2} \sqrt{x}}{\sqrt{x}} dx$.
Step $1$: Use substitution $\sqrt{x} = t$. Then $\frac{1}{2\sqrt{x}} dx = dt$,which implies $\frac{dx}{\sqrt{x}} = 2 dt$.
Step $2$: Substitute into the integral:
$I = \int \tan^4 t \cdot \sec^2 t \cdot (2 dt) = 2 \int \tan^4 t \sec^2 t dt$.
Step $3$: Use another substitution $\tan t = u$. Then $\sec^2 t dt = du$.
Step $4$: Integrate with respect to $u$:
$I = 2 \int u^4 du = 2 \cdot \frac{u^5}{5} + C = \frac{2}{5} u^5 + C$.
Step $5$: Substitute back $u = \tan t$ and $t = \sqrt{x}$:
$I = \frac{2}{5} \tan^5 \sqrt{x} + C$.