Question:
The value of \(∫^1_0tan^{-1}\frac{(2x-1)}{(1+x-x^2)}dx\) is
The value of \(∫^1_0tan^{-1}\frac{(2x-1)}{(1+x-x^2)}dx\) is
Updated On: Sep 18, 2023
1
0
-1
\(\frac{π}{4}\)
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The Correct Option is B
Solution and Explanation
The correct answer is B:0
Let \(I=∫^1_0tan^{-1}\frac{(2x-1)}{(1+x-x^2)}dx\)
\(⇒I=∫^1_0tan^{-1}\frac{(x-(1-x)}{(1+x(1-x))}dx\)
\(⇒I=∫^1_0[tan^{-1}x-tan^{-1}(1-x)]dx...(1)\)
\(⇒I=∫^1_0[tan^{-1}(1-x)-tan^{-1}(1-1+x)]dx\)
\(⇒I=∫^1_0[tan^{-1}(1-x)-tan^{-1}(x)]dx\)
\(⇒I=∫^1_0[tan^{-1}(1-x)-tan^{-1}(x)]dx...(2)\)
Adding(1)and(2),we obtain
\(2I=∫^1_0(tan^{-1}x+tan^{-1}(1-x)-tan^{-1}(1-x)-tan^{-1}x)dx\)
\(⇒2I=0\)
\(⇒I=0\)
Hence,the correct Answer is B.
Let \(I=∫^1_0tan^{-1}\frac{(2x-1)}{(1+x-x^2)}dx\)
\(⇒I=∫^1_0tan^{-1}\frac{(x-(1-x)}{(1+x(1-x))}dx\)
\(⇒I=∫^1_0[tan^{-1}x-tan^{-1}(1-x)]dx...(1)\)
\(⇒I=∫^1_0[tan^{-1}(1-x)-tan^{-1}(1-1+x)]dx\)
\(⇒I=∫^1_0[tan^{-1}(1-x)-tan^{-1}(x)]dx\)
\(⇒I=∫^1_0[tan^{-1}(1-x)-tan^{-1}(x)]dx...(2)\)
Adding(1)and(2),we obtain
\(2I=∫^1_0(tan^{-1}x+tan^{-1}(1-x)-tan^{-1}(1-x)-tan^{-1}x)dx\)
\(⇒2I=0\)
\(⇒I=0\)
Hence,the correct Answer is B.
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Concepts Used:
Definite Integral
Definite integral is an operation on functions which approximates the sum of the values (of the function) weighted by the length (or measure) of the intervals for which the function takes that value.
Definite integrals - Important Formulae Handbook
A real valued function being evaluated (integrated) over the closed interval [a, b] is written as :
\(\int_{a}^{b}f(x)dx\)
Definite integrals have a lot of applications. Its main application is that it is used to find out the area under the curve of a function, as shown below:
