Definite Integration Question 11
Question: The value of $ \int_0^{1}{{{\tan }^{-1}}( \frac{2x-1}{1+x-x^{2}} )}dx $ is
Options:
A) 1
B) 0
C) $ -1 $
D) None of these
Show Answer
Answer:
Correct Answer: B
Solution:
$ I=\int_0^{1}{{{\tan }^{-1}}}( \frac{2x-1}{1+x-x^{2}} )dx $
$ =\int_0^{1}{{{\tan }^{-1}}}( \frac{x+(x-1)}{1-x(x-1)} )dx $
$ I=\int_0^{1}{({{\tan }^{-1}}x+{{\tan }^{-1}}(x-1))}dx $
$ I=\int_0^{1}{{{\tan }^{-1}}xdx+\int_0^{1}{{{\tan }^{-1}}(x-1)dx}} $
$ I=\int_0^{1}{{{\tan }^{-1}}xdx+\int_0^{1}{{{\tan }^{-1}}(1-x-1)dx}} $ , {Using $ \int_0^{a}{f(x)dx=\int_0^{a}{f(a-x)dx}} $ in second integral} $ I=\int_0^{1}{{{\tan }^{-1}}xdx+\int_0^{1}{{{\tan }^{-1}}(-x)dx}} $
$ I=\int_0^{1}{{{\tan }^{-1}}xdx-\int_0^{1}{{{\tan }^{-1}}xdx}=0} $ .
BETA
