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  • Choose the correct answers in Exercises 41 to 44. The value of is

Choose the correct answers in Exercises 41 to 44.

    Q44.    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)    \frac{\pi}{4}

Answers (1)

best_answer

\\Let\ I=\int_0^1\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )dx\\

\\tan^{-1}\left(\frac{2x-1}{1 +x -x^2} \right )\\ =tan^{-1}\left ( \frac{x-(1-x)}{1+x(1-x)} \right )\\ =tan^{-1}x-tan^{-1}(1-x)               as      tan^{-1}\left ( \frac{a-b}{1+ab} \right )=tan^{-1}a-tan^{-1}b

Now the integral can be written as

\\I=\int_{0}^{1} \left ( tan^{-1}x-tan^{-1}(1-x) \right )dx\\ I=\int_{0}^{1} \left ( tan^{-1}(1-x)-tan^{-1}(1-(1-x)) \right )dx\\ I=\int_{0}^{1} \left ( tan^{-1}(1-x)-tan^{-1}x\right )dx\\ I=-I\\ 2I=0\\ I=0

(B) is correct.

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Sayak

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