The value of \( \int_{0}^{1}\displaystyle\frac{e^{x}}{1+e^{2x}}\,dx \) is
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \( \tan^{-1}e-\displaystyle\frac{\pi}{4} \)
Let \( t=e^{x} \), so \( dt=e^{x}\,dx \) and \( e^{2x}=t^2 \). Change the limits: x = 0 gives t = 1 and x = 1 gives t = e.
\( \displaystyle \int_{0}^{1}\frac{e^{x}}{1+e^{2x}}\,dx=\int_{1}^{e}\frac{dt}{1+t^2} \)
\( \displaystyle =\Big[\tan^{-1}t\Big]_{1}^{e} \)
\( \displaystyle =\tan^{-1}e-\tan^{-1}1 \)
\( \displaystyle =\tan^{-1}e-\frac{\pi}{4} \)