The value of \(\int_{0}^{1}\displaystyle\frac{dx}{e^{x}+e^{-x}}\) is
Answer & explanation
Correct answer: option 3
The correct answer is Option (3) → \(\tan^{-1}e-\displaystyle\frac{\pi}{4}\)
Multiplying the numerator and the denominator by \(e^{x}\):
\( \displaystyle \frac{1}{e^{x}+e^{-x}} = \frac{e^{x}}{e^{2x}+1} \)
Let \(t = e^{x}\). Then \(dt = e^{x}\,dx\).
When \(x = 0\), \(t = 1\); when \(x = 1\), \(t = e\). So
\( \displaystyle \int_{0}^{1}\frac{e^{x}}{e^{2x}+1}\,dx = \int_{1}^{e}\frac{dt}{t^{2}+1} \)
\( \displaystyle = \left[\tan^{-1}t\right]_{1}^{e} = \tan^{-1}e - \tan^{-1}1 \)
\( \displaystyle = \tan^{-1}e - \frac{\pi}{4} \)
Why the other options are incorrect
Option (1) \(\tan^{-1}e\) — Incorrect. The lower limit of t is 1, not 0, so \(\tan^{-1}1 = \frac{\pi}{4}\) must be subtracted.
Option (2) \(\tan^{-1}e-\displaystyle\frac{\pi}{2}\) — Incorrect. \(\tan^{-1}1 = \frac{\pi}{4}\), not \(\frac{\pi}{2}\); this value is also negative, though the integrand is positive.
Option (4) \(\tan^{-1}e+\displaystyle\frac{\pi}{4}\) — Incorrect. The lower-limit value is subtracted, not added.