Evaluate: ∫ tan⁻¹√x dx

Integrating by parts with $u = \tan^{-1}\sqrt{x}$, $dv = dx$: $\int \tan^{-1}\sqrt{x}\,dx = x\tan^{-1}\sqrt{x} - \int x\cdot\frac{1}{1+x}\cdot\frac{1}{2\sqrt{x}}dx$ $= x\tan^{-1}\sqrt{x} - \frac{1}{2}\int \frac{\sqrt{x}}{1+x}dx$. Put $\sqrt{x} = t$, $dx = 2t\,dt$: $\frac{1}{2}\int\frac{t\cdot 2t}{1+t^2}dt = \int\left(1 - \frac{1}{1+t^2}\right)dt = t - \tan^{-1}t$. Hence $\int \tan^{-1}\sqrt{x}\,dx = x\tan^{-1}\sqrt{x} - \sqrt{x} + \tan^{-1}\sqrt{x} + C = (x+1)\tan^{-1}\sqrt{x} - \sqrt{x} + C$.