Integrate the function w.r.t. $x$: $\frac{\tan^4 \sqrt{x} \sec^2 \sqrt{x}}{\sqrt{x}}$

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

The correct answer is Option (2) → $\frac{2}{5} \tan^5 \sqrt{x} + C$

Derivative of $\sqrt{x}$ is $\frac{1}{2} x^{-\frac{1}{2}} = \frac{1}{2\sqrt{x}}$. Thus, we use the substitution $\sqrt{x} = t$ so that $\frac{1}{2\sqrt{x}} \, dx = dt$ giving $dx = 2t \, dt$.

Thus, $\int \frac{\tan^4 \sqrt{x} \sec^2 \sqrt{x}}{\sqrt{x}} \, dx = \int \frac{2t \tan^4 t \sec^2 t \, dt}{t} = 2 \int \tan^4 t \sec^2 t \, dt$

Again, we make another substitution $\tan t = u$ so that $\sec^2 t \, dt = du$.

Therefore, $2 \int \tan^4 t \sec^2 t \, dt = 2 \int u^4 \, du = 2 \frac{u^5}{5} + C$

$= \frac{2}{5} \tan^5 t + C$ (since $u=\tan t$)

$= \frac{2}{5} \tan^5 \sqrt{x} + C$ (since $t=\sqrt{x}$)

Hence, $\frac{\tan^4 \sqrt{x} \sec^2 \sqrt{x}}{\sqrt{x}}dx=\frac{2}{5} \tan^5 \sqrt{x} + C$