Definite Integration Question 84
Question: $ \int _{0}^{1}{{{\tan }^{-1}}( \frac{1}{x^{2}-x+1} )dx} $ is
[Orissa JEE 2003]
Options:
A) ln 2
B) $ -\ln 2 $
C) $ \frac{\pi }{2}+\ln 2 $
D) $ \frac{\pi }{2}-\ln 2 $
Show Answer
Answer:
Correct Answer: D
Solution:
$ \int_0^{1}{{{\tan }^{-1}}( \frac{1}{x^{2}-x+1} )dx}=\int_0^{1}{{{\tan }^{-1}}xdx-}\int_0^{1}{{{\tan }^{-1}}(x-1)}dx $
$ =2\int _{0}^{1}{{{\tan }^{-1}}xdx}=2[{{\tan }^{-1}}x-\frac{1}{2}\log (1+x^{2})]_0^{1}=\frac{\pi }{2}-\log 2. $
BETA
