A) \[2{{\tan }^{-1}}x\]
B) \[2{{\tan }^{-1}}\sqrt{x}\]
C) \[2{{\cot }^{-1}}\sqrt{x}\]
D) \[{{\log }_{e}}(1+x)\]
Correct Answer: B
Solution :
\[I=\int_{{}}^{{}}{\frac{dx}{\sqrt{x}(1+{{(\sqrt{x})}^{2}})}}\]. Put \[\sqrt{x}=t\Rightarrow \frac{1}{2\sqrt{x}}\,dx=dt\] \[I=\int_{{}}^{{}}{\frac{2\,dt}{1+{{t}^{2}}}=2{{\tan }^{-1}}t+A}\] \[\therefore \,\,\,I=2{{\tan }^{-1}}\sqrt{x}+A\]; \[\therefore \,\,\,f(x)=2{{\tan }^{-1}}\sqrt{x}\].You need to login to perform this action.
You will be redirected in
3 sec