∫ tan^-1√x dx equals to: (Here C is an arbitrary constant)

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

Let \(x = t^2\), so that \(\sqrt{x} = t\) and \(dx = 2t\,dt\):

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

By integration by parts, with \(\tan^{-1}t\) as the first function:

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

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

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

Putting back \(t = \sqrt{x}\):

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

Shortcut: In the integration by parts, take the integral of 2t as \(t^2 + 1\) instead of \(t^2\) (any constant may be added). Then

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