Integral Calculus Question 480
Question: $ \int_{{}}^{{}}{\frac{dx}{2\sqrt{x}(1+x)}=} $
[RPET 2002]
Options:
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
Show Answer
Answer:
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 $ .
BETA
