Definite Integration Question 391
Question: The integral $ \int _{-1}^{3}{( {{\tan }^{-1}}\frac{x}{x^{2}+1}+{{\tan }^{-1}}\frac{x^{2}+1}{x} )}dx= $
[Karnataka CET 2000]
Options:
A) $ \pi $
B) $ 2\pi $
C) $ 3\pi $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int _{-1}^{3}{{ {{\tan }^{-1}}( \frac{x}{x^{2}+1} )+{{\tan }^{-1}}( \frac{x^{2}+1}{x} ) }}dx $
$ =\int _{-1}^{3}{{ {{\tan }^{-1}}( \frac{x}{x^{2}+1} )+{{\cot }^{-1}}( \frac{x}{x^{2}+1} ) }}dx $
$ =\int _{-1}^{3}{\frac{\pi }{2}dx=}2\pi $ .
BETA
