A) \[\frac{1}{2}{{\tan }^{-1}}(\sqrt{x})+c\]
B) \[{{\tan }^{-1}}(\sqrt{x})+c\]
C) \[2{{\tan }^{-1}}(\sqrt{x})+c\]
D) None of these
Correct Answer: B
Solution :
\[I=\int{\frac{dx}{2\sqrt{x}(1+x)}}\]. Put \[\sqrt{x}\,=t\]Þ \[\frac{1}{2\sqrt{x}}dx=dt\] \[\therefore I=\int{\frac{dt}{1+{{t}^{2}}}}={{\tan }^{-1}}t+c\]\[={{\tan }^{-1}}(\sqrt{x})+c\].
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