Integral Calculus Question 446
Question: If $ \int_{{}}^{{}}{\frac{1}{(1+x)\sqrt{x}}\ dx=f(x)+A} $ , where A is any arbitrary constant, then the function $ f(x) $ is
[MP PET 1998]
Options:
A) $ 2{{\tan }^{-1}}x $
B) $ 2{{\tan }^{-1}}\sqrt{x} $
C) $ 2{{\cot }^{-1}}\sqrt{x} $
D) $ {\log_{e}}(1+x) $
Show Answer
Answer:
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} $ .
BETA
