What is the integral of 1/(2+e^x) ?

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(2+e^x)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{1}{u ( 2 + u )} d u

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

Take the partial fraction of

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

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

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

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

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

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

Substitute back

u = e^{x}

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

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

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

Add a constant to the solution

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

\int t + 6 d t

f (x) = \frac{3 x ^{5} - 7 x ^{2} - 4}{x ^{2}}

x \to 0 lim (cot^{3} (x))

\frac{d}{d x} (\frac{tan ( 20 x )}{x ^{\frac{2}{3}} + 3 x})

x \to 0 lim (\frac{x}{ln ( x + 1 )})

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