Definite Integration Question 275
Question: $ \int _{0}^{\infty }{\frac{xdx}{(1+x)(1+x^{2})}=} $
[Karnataka CET 2003]
Options:
A) 0
B) $ \pi /2 $
C) $ \pi /4 $
D) 1
Show Answer
Answer:
Correct Answer: C
Solution:
$ \int_0^{\infty }{\frac{xdx}{(1+x)(1+x^{2})}=\int_0^{\infty }{\frac{-\frac{1}{2}dx}{(1+x)}+\int_0^{\infty }{\frac{( \frac{1}{2}x+\frac{1}{2} )}{1+x^{2}}dx}}} $
$ =[ \frac{-1}{2}\log (1+x) ]_0^{\infty }+\frac{1}{2}\times \frac{1}{2}[\log (1+x^{2})] _0^{\infty }+\frac{1}{2}[{{\tan }^{-1}}x] _0^{\infty } $
$ =0+0+\frac{1}{2}[ \frac{\pi }{2}-0 ]=\frac{\pi }{4} $ .
BETA
