What is the integral of (e^x)/(e^x-2e^{-x)} ?

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{u ^{2} - 2} d u

= \int \frac{u}{u ^{2} - 2} d u

Apply u-substitution

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

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

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}{v} d v

Use the common integral:

\int \frac{1}{v} d v = ln (∣ v ∣)

= \frac{1}{2} ln ∣ v ∣

Substitute back

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

Apply exponent rule:

(a^{b})^{c} = a^{b c}

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

Add a constant to the solution

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

\frac{x ^{2}}{( x ^{2} + 5 ) ^{3}}

y^{^{'}} = \frac{7 x}{y}

\int \frac{x ^{2}}{8 x ^{3} - 1} d x

(2.4) (1.6)

f (x) = \frac{2}{x - 1}, at x = 3

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