What is the integral of 1/(e^{2x)+4e^x} ?

The integral of 1/(e^{2x)+4e^x} is -1/16 x-1/(4e^x)+1/16 ln|e^x+4|+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of 1/(e^{2x)+4e^x}

\int \frac{1}{e ^{2 x} + 4 e ^{x}} d x

- \frac{1}{16} x - \frac{1}{4 e ^{x}} + \frac{1}{16} ln ∣ e^{x} + 4 ∣ + C

Solution steps

\int \frac{1}{e ^{2 x} + 4 e ^{x}} d x

Apply u-substitution

: \int \frac{1}{u ( u ^{2} + 4 u )} d u

= \int \frac{1}{u ( u ^{2} + 4 u )} d u

Take the partial fraction of

\frac{1}{u ( u ^{2} + 4 u )} : - \frac{1}{16 u} + \frac{1}{4 u ^{2}} + \frac{1}{16 ( u + 4 )}

= \int - \frac{1}{16 u} + \frac{1}{4 u ^{2}} + \frac{1}{16 ( u + 4 )} 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}{16 u} d u + \int \frac{1}{4 u ^{2}} d u + \int \frac{1}{16 ( u + 4 )} d u

\int \frac{1}{16 u} d u = \frac{1}{16} ln ∣ u ∣

\int \frac{1}{4 u ^{2}} d u = - \frac{1}{4 u}

\int \frac{1}{16 ( u + 4 )} d u = \frac{1}{16} ln ∣ u + 4 ∣

= - \frac{1}{16} ln ∣ u ∣ - \frac{1}{4 u} + \frac{1}{16} ln ∣ u + 4 ∣

Substitute back

u = e^{x}

= - \frac{1}{16} ln ∣ e^{x} ∣ - \frac{1}{4 e ^{x}} + \frac{1}{16} ln ∣ e^{x} + 4 ∣

\frac{1}{16} ln ∣ e^{x} ∣ = \frac{1}{16} x

= - \frac{1}{16} x - \frac{1}{4 e ^{x}} + \frac{1}{16} ln ∣ e^{x} + 4 ∣

Add a constant to the solution

= - \frac{1}{16} x - \frac{1}{4 e ^{x}} + \frac{1}{16} ln ∣ e^{x} + 4 ∣ + C

\int sin (3 y) ln (x)

\int \frac{9 x}{x ^{3} - 3 x + 2} d x

\int \frac{1}{- x ^{2} + 10 x - 16} d x

\int \frac{1}{x ^{4} x ^{2} - 36} d x

\int \frac{e x}{n} d x

What is the integral of 1/(e^{2x)+4e^x} ?
The integral of 1/(e^{2x)+4e^x} is -1/16 x-1/(4e^x)+1/16 ln|e^x+4|+C