$\int\tan^{-1}\sqrt{x} dx$ equals to: (Here C is an arbitrary constant)

$(x+1)\tan^{-1}\sqrt{x}-\sqrt{x}+C$

The correct answer is Option (1) → $(x+1)\tan^{-1}\sqrt{x}-\sqrt{x}+C$

Evaluate: $\int \tan^{-1}(\sqrt{x}) \, dx$

Substitute: $t = \sqrt{x} = x^{1/2}$, so $x = t^2$ and $dx = 2t \, dt$

The integral becomes:

$\int \tan^{-1}(\sqrt{x}) \, dx = \int \tan^{-1}(t) \cdot 2t \, dt = 2 \int t \tan^{-1}(t) \, dt$

Use integration by parts:

Let:

$u = \tan^{-1}(t)$

$du = \frac{1}{1 + t^2} dt$

$dv = t \, dt$

$v = \frac{t^2}{2}$

Then,

$2 \int t \tan^{-1}(t) \, dt = 2 \left( u v - \int v \, du \right) = 2 \left( \frac{t^2}{2} \tan^{-1}(t) - \int \frac{t^2}{2} \cdot \frac{1}{1 + t^2} dt \right)$

$= t^2 \tan^{-1}(t) - \int \frac{t^2}{1 + t^2} dt$

Simplify the integral:

$\int \frac{t^2}{1 + t^2} dt = \int \left(1 - \frac{1}{1 + t^2}\right) dt = \int 1 \, dt - \int \frac{1}{1 + t^2} dt = t - \tan^{-1}(t) + C$

Therefore, the original integral is:

$t^2 \tan^{-1}(t) - (t - \tan^{-1}(t)) + C = t^2 \tan^{-1}(t) - t + \tan^{-1}(t) + C$

Substitute back $t = \sqrt{x}$:

${\int \tan^{-1}(\sqrt{x}) \, dx = x \tan^{-1}(\sqrt{x}) - \sqrt{x} + \tan^{-1}(\sqrt{x}) + C}$