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KCET · Maths · Definite Integration

\( \int_{0}^{1} \frac{d x}{e^{x}+e^{-x}} \) is equal to

  1. A \( \frac{\pi}{4}-\tan ^{-1}(\mathrm{e}) \)
  2. B \( \tan ^{-1}(e)-\frac{\pi}{4} \)
  3. C \( \tan ^{-1}(e)+\frac{\pi}{4} \)
  4. D \( \tan ^{-1}(\mathrm{e}) \)
Verified Solution

Answer & Solution

Correct Answer

(B) \( \tan ^{-1}(e)-\frac{\pi}{4} \)

Step-by-step Solution

Detailed explanation

Given that \( \int_{0}^{1} \frac{d x}{e^{x}+e^{-x}}=\int_{0}^{1} \frac{e^{x}}{e^{2 x}+1} d x \)
\( =\left[\tan ^{-1}\left(e^{x}\right)\right]_{0}^{1}=\tan ^{-1}(e)-\frac{\pi}{4} \)