1-1-e-x-e-x-dx-

Question Number 143506 by tugu last updated on 15/Jun/21

\underset{1}{\int^{\infty}} \frac{1}{e^{- x} + e^{x}} d x = ?

Answered by Dwaipayan Shikari last updated on 15/Jun/21

\int_{1}^{\infty} \frac{1}{e^{x} + e^{- x}} d x e^{x} = t

= \int_{e}^{\infty} \frac{d t}{t^{2} + 1} = {[{t a n}^{- 1} (t)]}_{e}^{\infty} = \frac{π}{2} - {t a n}^{- 1} (e) = {t a n}^{- 1} (\frac{1}{e}) = 0.3525