Integral Calculus Question 112
Question: $ \int_{{}}^{{}}{\frac{x{{\tan }^{-1}}x}{{{(1+x^{2})}^{3/2}}}\ dx=} $
Options:
A) $ \frac{x+{{\tan }^{-1}}x}{\sqrt{1+x^{2}}}+c $
B) $ \frac{x-{{\tan }^{-1}}x}{\sqrt{1+x^{2}}}+c $
C) $ \frac{{{\tan }^{-1}}x-x}{\sqrt{1+x^{2}}}+c $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
Put $ x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta ,d\theta , $ then $ \int_{{}}^{{}}{\frac{x{{\tan }^{-1}}x}{{{(1+x^{2})}^{3/2}}},dx}=\int_{{}}^{{}}{\frac{\theta \tan \theta {{\sec }^{2}}\theta ,d\theta }{{{(1+{{\tan }^{2}}\theta )}^{3/2}}}} $ $ =\int_{{}}^{{}}{\theta \sin \theta ,d\theta }=-\theta \cos \theta +\sin \theta +c $ $ =\frac{x}{\sqrt{x^{2}+1}}-{{\tan }^{-1}}x\frac{1}{\sqrt{x^{2}+1}}=\frac{x-{{\tan }^{-1}}x}{\sqrt{1+x^{2}}}+c $ .
BETA
