Q. The value of $\int_{0}^{1}{\tan^{- 1}{\left( \frac{2x - 1}{1 + x - x^{2}} \right)dx}}$ is
A) 1
B) 0
C) $- 1$
D) None of these
Explanation
(b)
Let $I = \int_{0}^{1}{\tan^{- 1}{\left( \frac{2x - 1}{1 + x - x^{2}} \right)dx}}$
$= \int_{0}^{1}{\tan^{- 1}{x\ dx + \int_{0}^{1}{\tan^{- 1}{(x - 1)dx}}}}$
$= \int_{0}^{1}{\tan^{- 1}{x\ dx + \int_{0}^{1}{\tan^{- 1}{(1 - x - 1)dx}}}}$
$= \int_{0}^{1}{\tan^{- 1}{x\ dx - \int_{0}^{1}{{\tan^{- 1}x}{dx = 0}}}}$
Reviewed by Dr. Abubacker
Physics Educator & NEET/JEE Specialist
Ph.D. in Physics with 19+ years of teaching experience.