Mathematics - Indefinite Integration Question with Solution | TestHub

We have, $I=\int 1\cdot {\tan }^{-1}\sqrt{\mathrm{x}}\mathrm{dx}$
Using by parts,
$\begin{array}{rl}I& ={\tan }^{-1}\sqrt{x}\cdot (x)-\int \frac{1}{1+x}\times \frac{1}{2\sqrt{x}}\times xdx\\ & =x{\tan }^{-1}\sqrt{x}\cdot -\int \frac{x}{(1+x)2\sqrt{x}}dx\\ & =x{\tan }^{-1}\sqrt{x}-\int \left(\frac{1+x}{(1+x)2\sqrt{x}}-\frac{1}{(1+x)2\sqrt{x}}\right)dx\\ & =x{\tan }^{-1}\sqrt{x}-\int \frac{dx}{2\sqrt{x}}+\int \frac{dx}{2\sqrt{x}(1+x)}\\ & =x{\tan }^{-1}\sqrt{x}-\sqrt{x}+{\tan }^{-1}\sqrt{x}+C\\ & =(x+1){\tan }^{-1}\sqrt{x}-\sqrt{x}+C\end{array}$