KCET · Maths · Definite Integration
\( \int_{0}^{1} \frac{d x}{e^{x}+e^{-x}} \) is equal to
- A \( \frac{\pi}{4}-\tan ^{-1}(\mathrm{e}) \)
- B \( \tan ^{-1}(e)-\frac{\pi}{4} \)
- C \( \tan ^{-1}(e)+\frac{\pi}{4} \)
- D \( \tan ^{-1}(\mathrm{e}) \)
Answer & Solution
Correct Answer
(B) \( \tan ^{-1}(e)-\frac{\pi}{4} \)
Step-by-step Solution
Detailed explanation
Given that \( \int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}=\int_{0}^{1} \frac{e^{x}}{e^{2 x}+1} d x \)
\( =\left[\tan ^{-1}\left(e^{x}\right)\right]_{0}^{1}=\tan ^{-1}(e)-\frac{\pi}{4} \)
\( =\left[\tan ^{-1}\left(e^{x}\right)\right]_{0}^{1}=\tan ^{-1}(e)-\frac{\pi}{4} \)
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