Question

If $I=\int {\tan }^{-1}(1+\sqrt{x})dx$, then $I$ equals

• A. ${(\sqrt{x}-1)}^2{\tan }^{-1}(1+\sqrt{x})-\sqrt{x}+\log (x+2\sqrt{x}+2)+C$
• B. ${(2\sqrt{2}+1)}^2{\tan }^{-1}(1+\sqrt{x})+\sqrt{x}+\log (x+2\sqrt{x}+2)+C$
• C. ${\left(\sqrt{x-1}\right)}^2{\tan }^{-1}(1+\sqrt{x})+\sqrt{x}+\log (\sqrt{x}+1)+C$
• D. None of this

Answer 1

The correct option is D None of this

$I=\int {\tan }^{-1}\left(1+\sqrt{x}\right)dxLet,\therefore \left(1+\sqrt{x}\right)=y\therefore \frac{1}{2\left(1+\sqrt{x}\right)}dx=dy\therefore dx=2\left(1+\sqrt{x}\right)dy\therefore dx=2ydy\therefore \int {\tan }^{-1}y\left(2ydy\right)\therefore 2\int y{\tan }^{-1}ydy\therefore 2\left[\frac{2y^2-1}{4}{\tan }^{-1}y+\frac{y\sqrt{1-y^2}}{4}\right]+c\therefore \frac{2y^2-1}{2}{\tan }^{-1}y+\frac{y\sqrt{1-y^2}}{2}+c\therefore 2\frac{{\left(1+\sqrt{x}\right)}^2-1}{2}{\tan }^{-1}\left(1+\sqrt{x}\right)+\frac{\left(1+\sqrt{x}\right)\sqrt{1-\left(1+\sqrt{x^2}\right)}}{}+c\therefore \frac{2{\left(1+\sqrt{x}\right)}^2-1}{2}{\tan }^{-1}\left(1+\sqrt{x}\right)+\frac{\left(1+\sqrt{x}\right)\left(1-\sqrt{x}\right)}{2}+c$

Hence,

None of these