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Question: \[\int_{}^{}{\frac{\mathbf{dx}}{\mathbf{2}\sqrt{\mathbf{x}}\mathbf{(1 + x)}}\mathbf{=}}\]...

∫dx2x(1+x)=\int_{}^{}{\frac{\mathbf{dx}}{\mathbf{2}\sqrt{\mathbf{x}}\mathbf{(1 + x)}}\mathbf{=}}

A

12tan⁡−1(x)+c\frac{1}{2}\tan^{- 1}(\sqrt{x}) + c

B

tan⁡−1(x)+C\tan^{- 1}(\sqrt{x}) + C

C

2tan⁡−1(x)+C2\tan^{- 1}(\sqrt{x}) + C

D

None of these

Answer

tan⁡−1(x)+C\tan^{- 1}(\sqrt{x}) + C

Explanation

Solution

I=∫dx2x(1+x),I = \int_{}^{}\frac{dx}{2\sqrt{x}(1 + x)}, put x=t\sqrt{x} = t ⇒12xdx=dt\Rightarrow \frac{1}{2\sqrt{x}}dx = dt

∴I=∫dt1+t2=tan⁡−1t+c=tan⁡−1x+c\therefore I = \int_{}^{}\frac{dt}{1 + t^{2}} = \tan^{- 1}t + c = \tan^{- 1}\sqrt{x} + c