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

The integral to find is $\int \tan^{-1}\sqrt{x} \, dx$

Using integration by parts with:

$u = \tan^{-1}\sqrt{x}$ and $dv = dx$

Then $v = x$

For $du$, using the chain rule:

$du = \frac{1}{1+(\sqrt{x})^2} \cdot \frac{1}{2\sqrt{x}} \, dx$

$du = \frac{1}{2\sqrt{x}(1+x)} \, dx$

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Applying integration by parts:

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

$= x \tan^{-1}\sqrt{x} - \int \frac{x}{2\sqrt{x}(1+x)} \, dx$

$= x \tan^{-1}\sqrt{x} - \int \frac{\sqrt{x}}{2(1+x)} \, dx$

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For $\int \frac{\sqrt{x}}{2(1+x)} \, dx$, let $t = \sqrt{x}$

Then $x = t^2$ and $dx = 2t \, dt$

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

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

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Rewriting the integrand:

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

$= \frac{t^2+1}{1+t^2} - \frac{1}{1+t^2}$

$= 1 - \frac{1}{1+t^2}$

Therefore:

$\int \frac{t^2}{1+t^2} \, dt = \int 1 \, dt - \int \frac{1}{1+t^2} \, dt$

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

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

$= \sqrt{x} - \tan^{-1}\sqrt{x}$

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Combining all parts:

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

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

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

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