The value of _-1^3 [^-1(x/x^2+1) + ^-1(x^2+1/x)] dx is: - Mathematics MCQ Mathematics Mock Test - 3 2026 | ExamTest.live

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The value of \int_{- 1}^{3} [tan^{- 1} (\frac{x}{x ^{2} + 1}) + tan^{- 1} (\frac{x ^{2} + 1}{x})] d x is:

Solution & Step-by-step Explanation

Use the identity tan^{- 1} (p) + tan^{- 1} (1/ p) = π /2 for p > 0 .Here p = \frac{x}{x ^{2} + 1} .The integral simplifies to: \int_{- 1}^{3} \frac{π}{2} d x = \frac{π}{2} [x]_{- 1}^{3} \frac{π}{2} (3 - (- 1)) = \frac{π}{2} \times 4 = 2 π

Try it yourself before checking the explanation above.

The value of \int_{- 1}^{3} [tan^{- 1} (\frac{x}{x ^{2} + 1}) + tan^{- 1} (\frac{x ^{2} + 1}{x})] d x is:

π

π /4

π /2

2 π

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