What is the integral of 3/(1+2e^x-e^{-x)} ?

The integral of 3/(1+2e^x-e^{-x)} is ln|2e^x-1|-ln|e^x+1|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 3/(1+2e^x-e^{-x)}

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

ln ∣ 2 e^{x} - 1 ∣ - ln ∣ e^{x} + 1 ∣ + C

Solution steps

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

Take the constant out:

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

= 3 \cdot \int \frac{1}{1 + 2 e ^{x} - e ^{- x}} d x

Apply u-substitution

: \int \frac{1}{u + 2 u ^{2} - 1} d u

= 3 \cdot \int \frac{1}{u + 2 u ^{2} - 1} d u

Take the partial fraction of

\frac{1}{u + 2 u ^{2} - 1} : \frac{2}{3 ( 2 u - 1 )} - \frac{1}{3 ( u + 1 )}

= 3 \cdot \int \frac{2}{3 ( 2 u - 1 )} - \frac{1}{3 ( u + 1 )} d u

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= 3 (\int \frac{2}{3 ( 2 u - 1 )} d u - \int \frac{1}{3 ( u + 1 )} d u)

\int \frac{2}{3 ( 2 u - 1 )} d u = \frac{1}{3} ln ∣ 2 u - 1 ∣

\int \frac{1}{3 ( u + 1 )} d u = \frac{1}{3} ln ∣ u + 1 ∣

= 3 (\frac{1}{3} ln ∣ 2 u - 1 ∣ - \frac{1}{3} ln ∣ u + 1 ∣)

Substitute back

u = e^{x}

= 3 (\frac{1}{3} ln ∣ 2 e^{x} - 1 ∣ - \frac{1}{3} ln ∣ e^{x} + 1 ∣)

Simplify

3 (\frac{1}{3} ln ∣ 2 e^{x} - 1 ∣ - \frac{1}{3} ln ∣ e^{x} + 1 ∣) : ln ∣ 2 e^{x} - 1 ∣ - ln ∣ e^{x} + 1 ∣

= ln ∣ 2 e^{x} - 1 ∣ - ln ∣ e^{x} + 1 ∣

Add a constant to the solution

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

t \to - 1 lim (t^{2} - x^{2})

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

\frac{\partial}{\partial z} (z - x)

\frac{d}{d x} (\frac{1}{x ^{5}} + \frac{1}{x ^{6}})

\int (x + 1) 6 - x d x

What is the integral of 3/(1+2e^x-e^{-x)} ?
The integral of 3/(1+2e^x-e^{-x)} is ln|2e^x-1|-ln|e^x+1|+C