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Solve the following.
Evaluate: ∫ tan⁻¹√x dx.
Evaluate: $\int \tan^{-1}\sqrt{x} \, dx$.
Answer
Answer
Integral: $\int \left(\tan^{-1}(\sqrt{x})\right) dx$
Put $x = t^2$, so $dx = 2t\,dt$ and $\sqrt{x} = t$.
$I = \int 2t\tan^{-1}t\,dt = t^2\tan^{-1}t - \int \frac{t^2}{1+t^2}dt$ (integration by parts)
$= t^2\tan^{-1}t - \int\left(1 - \frac{1}{1+t^2}\right)dt = t^2\tan^{-1}t - t + \tan^{-1}t + C$
Hence $I = (x+1)\tan^{-1}\sqrt{x} - \sqrt{x} + C$.
Final answer: $(x+1)\tan^{-1}\sqrt{x} - \sqrt{x}$