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

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

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

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

Solution steps

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

Apply u-substitution

: \int \frac{u}{u ^{2} + 3 u + 2} d u

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

Take the partial fraction of

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

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

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

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

= - ln ∣ u + 1 ∣ + 2 ln ∣ u + 2 ∣

Substitute back

u = e^{x}

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

Add a constant to the solution

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

\frac{\partial}{\partial x} (tan (ln (x - y)))

\frac{\partial}{\partial x} (\frac{e ^{x y}}{x + y})

\int (2 π cos (π x)) d x

\int - sin (x) cos (x) d x

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