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Question

Mathematics Question on Integrals of Some Particular Functions

∫01x tan⁡−1x dx=\int^1_0 x \, \tan^{-1} x\,dx =

A

π4+12\frac{\pi}{4} + \frac{1}{2}

B

π4−12\frac{\pi}{4} - \frac{1}{2}

C

12−π4\frac{1}{2} - \frac{\pi}{4}

D

−π4−12 - \frac{\pi}{4} - \frac{1}{2}

Answer

π4−12\frac{\pi}{4} - \frac{1}{2}

Explanation

Solution

Answer (b) π4−12\frac{\pi}{4} - \frac{1}{2}