Question:
\(\int^{1}_{0}\frac{2e^x}{1+e^{2x}}dx=\)
\(\int^{1}_{0}\frac{2e^x}{1+e^{2x}}dx=\)
Updated On: Jun 10, 2024
- \(4(\tan^{-1}2-\pi)\)
- \(2(\tan^{-1}e-\frac{\pi}{2})\)
- \(2(\tan^{-1}e+\frac{\pi}{4})\)
- \(2(\tan^{-1}e-\frac{\pi}{4})\)
- \(2(\tan^{-1}2+\pi)\)
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The Correct Option is D
Solution and Explanation
The correct option is (D): \(2(\tan^{-1}e-\frac{\pi}{4})\)
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