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Question

Mathematics Question on Integration

∫012ex1+e2xdx=\int^{1}_{0}\frac{2e^x}{1+e^{2x}}dx=

A

4(tan⁡−12−π)4(\tan^{-1}2-\pi)

B

2(tan⁡−1e−π2)2(\tan^{-1}e-\frac{\pi}{2})

C

2(tan⁡−1e+π4)2(\tan^{-1}e+\frac{\pi}{4})

D

2(tan⁡−1e−π4)2(\tan^{-1}e-\frac{\pi}{4})

E

2(tan⁡−12+π)2(\tan^{-1}2+\pi)

Answer

2(tan⁡−1e−π4)2(\tan^{-1}e-\frac{\pi}{4})

Explanation

Solution

The correct option is (D): 2(tan⁡−1e−π4)2(\tan^{-1}e-\frac{\pi}{4})