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Mathematics Question on integral

The value of ∫01tan−1(2x−1)(1+x−x2)dx∫^1_0tan^{-1}\frac{(2x-1)}{(1+x-x^2)}dx is

A

1

B

0

C

-1

D

π4\frac{π}{4}

Answer

0

Explanation

Solution

The correct answer is B:0
Let I=∫01tan−1(2x−1)(1+x−x2)dxI=∫^1_0tan^{-1}\frac{(2x-1)}{(1+x-x^2)}dx
⇒I=∫01tan−1(x−(1−x)(1+x(1−x))dx⇒I=∫^1_0tan^{-1}\frac{(x-(1-x)}{(1+x(1-x))}dx
⇒I=∫01[tan−1x−tan−1(1−x)]dx...(1)⇒I=∫^1_0[tan^{-1}x-tan^{-1}(1-x)]dx...(1)
⇒I=∫01[tan−1(1−x)−tan−1(1−1+x)]dx⇒I=∫^1_0[tan^{-1}(1-x)-tan^{-1}(1-1+x)]dx
⇒I=∫01[tan−1(1−x)−tan−1(x)]dx⇒I=∫^1_0[tan^{-1}(1-x)-tan^{-1}(x)]dx
⇒I=∫01[tan−1(1−x)−tan−1(x)]dx...(2)⇒I=∫^1_0[tan^{-1}(1-x)-tan^{-1}(x)]dx...(2)
Adding(1)and(2),we obtain
2I=∫01(tan−1x+tan−1(1−x)−tan−1(1−x)−tan−1x)dx2I=∫^1_0(tan^{-1}x+tan^{-1}(1-x)-tan^{-1}(1-x)-tan^{-1}x)dx
⇒2I=0⇒2I=0
⇒I=0⇒I=0
Hence,the correct Answer is B.