$\mathop \smallint \nolimits_0^1 \,\,{\tan ^{ - 1}}\,\,\left( {\frac{1}{{{x^2} - x + 1}}} \right)\,\,dx$ is equal to

$\frac{{\rm{\pi }}}{2}\,\, - \,\,\log \,\,2$

$\frac{{\rm{\pi }}}{2}\,\, + \,\,\log \,\,2$

JEE MATHEMATICS MCQ: DEFINITE INTEGRATION

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Application +4 | -1

\int \nolimits_0^1 \,\,{\tan ^{ - 1}}\,\,\left( {\frac{1}{{{x^2} - x + 1}}} \right)\,\,dx

is equal to

View answer and explanation

Correct option: D

\frac{\mathrm{\pi }}{2}\,\, - \,\,\log \,\,2

The correct answer is D (

\frac{\mathrm{\pi }}{2}\\ - \log \\,2

) because the integral can be simplified using trigonometric identities and integration by parts.

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