What is the integral of 1/(e^t+e^{-t)} ?

The integral of 1/(e^t+e^{-t)} is arctan(e^t)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of 1/(e^t+e^{-t)}

\int \frac{1}{e ^{t} + e ^{- t}} d t

arctan (e^{t}) + C

Solution steps

\int \frac{1}{e ^{t} + e ^{- t}} d t

Apply u-substitution

= \int \frac{1}{u ^{2} + 1} d u

Use the common integral:

\int \frac{1}{u ^{2} + 1} d u = arctan (u)

= arctan (u)

Substitute back

u = e^{t}

= arctan (e^{t})

Add a constant to the solution

= arctan (e^{t}) + C

What is the integral of 1/(e^t+e^{-t)} ?
The integral of 1/(e^t+e^{-t)} is arctan(e^t)+C