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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

integral

\frac{1}{( e ^{x} + e ^{- x} )}

arctan (e^{x}) + C

Solution steps

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

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^{x}

= arctan (e^{x})

Add a constant to the solution

= arctan (e^{x}) + C

in t e g r a l \frac{x}{( x + 2 )}

in t e g r a l (e^{- t})^{2}

in t e g r a l 2 x (x^{2} + 1)^{\frac{1}{2}}

in t e g r a l \frac{1}{- x}

in t e g r a l \frac{( y + 1 )}{( y - 1 )}

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