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

The integral of (e^x)/(e^x-e^{-x)} is 1/2 ln|e^{2x}-1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{2} ln e^{2 x} - 1 + C

Solution steps

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

Factor

\frac{e ^{x}}{e ^{x} - e ^{- x}}

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

Multiply by the conjugate of

e^{x} + 1 : \frac{e ^{2 x} - 1}{e ^{x} - 1}

= \int \frac{e ^{x}}{e ^{- x} \frac{e ^{2 x} - 1}{e ^{x} - 1} ( e ^{x} - 1 )} d x

Simplify

\frac{e ^{x}}{e ^{- x} \frac{e ^{2 x} - 1}{e ^{x} - 1} ( e ^{x} - 1 )} : \frac{e ^{2 x}}{e ^{2 x} - 1}

= \int \frac{e ^{2 x}}{e ^{2 x} - 1} d x

Apply u-substitution

: \int \frac{1}{2 u} d u

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

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{2} \cdot \int \frac{1}{u} d u

Use the common integral:

\int \frac{1}{u} d u = ln (∣ u ∣)

= \frac{1}{2} ln ∣ u ∣

Substitute back

u = e^{2 x} - 1

= \frac{1}{2} ln e^{2 x} - 1

Add a constant to the solution

= \frac{1}{2} ln e^{2 x} - 1 + C

\int (x + \frac{1}{x ^{3}} - 3 e^{3 x}) d x

\int \frac{1}{( x x ^{2} - 1 )} d x

\int (5 - 4 x) cos (6 - 5 x) d x

\int 7 x^{2} e^{x^{3}} d x

\int x e^{a x^{2}} d x

What is the integral of (e^x)/(e^x-e^{-x)} ?
The integral of (e^x)/(e^x-e^{-x)} is 1/2 ln|e^{2x}-1|+C