Definite-Integration Question 535
Question: $ \int_{,0}^{,\infty }{\frac{x\ln x,dx}{{{(1+x^{2})}^{2}}}} $ is equal to
[AMU 2000]
Options:
A) 0
B) 1
C) $ \infty $
D) None of these
Show Answer
Answer:
Correct Answer: A
Solution:
$ I=\int_0^{\infty }{\frac{x\log x}{{{(1+x^{2})}^{2}}},dx} $ Put $ x=\tan \theta $
Þ $ dx={{\sec }^{2}}\theta ,d\theta $
$ \therefore $ I $ =\int_0^{\pi /2}{\frac{\tan \theta ,\log ,(\tan \theta )}{{{\sec }^{4}}\theta }}{{\sec }^{2}}\theta ,d\theta $ $ =\int_0^{\pi /2}{\sin \theta ,\cos \theta ,\log ,(\tan \theta ),d\theta } $ $ =\frac{1}{2}\int_0^{\pi /2}{\sin 2\theta \log ,(,\tan \theta ,),d\theta } $ $ =0 $ , $ { \because \int_0^{\pi /2}{\sin 2\theta \log \tan \theta d\theta =0} } $ .
BETA
