A) \[{{\tan }^{-1}}\sqrt{x}+c\]
B) \[2{{\tan }^{-1}}\sqrt{x}+c\]
C) \[2{{\tan }^{-1}}x+c\]
D) None of these
Correct Answer: B
Solution :
Let \[I=\int{\frac{dx}{\sqrt{x}(1+x)}}\] Let \[x=ta{{n}^{2}}\theta \] \[\Rightarrow \] \[dx=2\text{ }tan\text{ }\theta .se{{c}^{2}}\theta d\theta \] \[\therefore \]\[I=\int{\frac{1}{\tan \theta (1+{{\tan }^{2}}\theta )}}2\tan \theta .{{\sec }^{2}}\theta .d\theta \] \[=\int{\frac{2\tan \theta .{{\sec }^{2}}\theta }{\tan \theta {{\sec }^{2}}\theta }}d\theta =\int{2d\theta }\] \[=2\theta +c=2{{\tan }^{-1}}\sqrt{x}+c\]
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