Integral Calculus Question 492
Question: $ \int_{{}}^{{}}{\frac{\sqrt{x}}{1+x}dx=} $
Options:
A) $ \sqrt{x}-{{\tan }^{-1}}\sqrt{x}+c $
B) $ 2(\sqrt{x}-{{\tan }^{-1}}\sqrt{x})+c $
C) $ 2(\sqrt{x}+{{\tan }^{-1}}x)+c $
D) $ \sqrt{1+x}+c $
Show Answer
Answer:
Correct Answer: B
Solution:
$ \int_{{}}^{{}}{\frac{\sqrt{x}}{1+x},dx=\int_{{}}^{{}}{\frac{\sqrt{x}\sqrt{x}}{\sqrt{x}(1+x)},dx}} $ $ =\int_{{}}^{{}}{\frac{x+1}{\sqrt{x}(x+1)},dx-\int_{{}}^{{}}{\frac{1}{\sqrt{x}(x+1)},dx}} $ $ =\int_{{}}^{{}}{\frac{1}{\sqrt{x}},dx-\int_{{}}^{{}}{\frac{1}{\sqrt{x}(x+1)},dx}} $ $ =2{x^{1/2}}-2{{\tan }^{-1}}\sqrt{x}+c=2(\sqrt{x}-{{\tan }^{-1}}\sqrt{x})+c. $
BETA
