Integral Calculus Question 408
Question: $ \int_{{}}^{{}}{{{\tan }^{-1}}\frac{2x}{1-x^{2}}dx=} $
[MP PET 1991]
Options:
A) $ x{{\tan }^{-1}}x+c $
B) $ x{{\tan }^{-1}}x-\log (1+x^{2})+c $
C) $ 2x{{\tan }^{-1}}x+\log (1+x^{2})+c $
D) $ 2x{{\tan }^{-1}}x-\log (1+x^{2})+c $
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Answer:
Correct Answer: D
Solution:
Put $ x=\tan \theta \Rightarrow dx={{\sec }^{2}}\theta ,d\theta , $ then $ \int_{{}}^{{}}{{{\tan }^{-1}}\frac{2x}{1-x^{2}},dx}=\int_{{}}^{{}}{{{\tan }^{-1}}\frac{2\tan \theta }{1-{{\tan }^{2}}\theta }},{{\sec }^{2}}\theta ,d\theta $ $ =\int_{{}}^{{}}{{{\tan }^{-1}}(\tan 2\theta ){{\sec }^{2}}\theta ,d\theta }=\int_{{}}^{{}}{2\theta {{\sec }^{2}}\theta ,d\theta } $ $ =2[ \theta \tan \theta -\int_{{}}^{{}}{\tan \theta ,d\theta } ] $ $ =2x{{\tan }^{-1}}x-\log (x^{2}+1)+c. $
BETA
