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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

\frac{1}{3} ln e^{3 x} + 1 + 2 (\frac{1}{3} ln ∣ e^{x} + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 e ^{2 x} - 4 e ^{x}}{3} + 23 arctan (\frac{2 e ^{x} - 1}{3}))) + C

Solution steps

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

Expand

\frac{e ^{3 x} + 2 e ^{x}}{e ^{3 x} + 1} : \frac{e ^{3 x}}{e ^{3 x} + 1} + \frac{2 e ^{x}}{e ^{3 x} + 1}

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

\int \frac{e ^{3 x}}{e ^{3 x} + 1} d x = \frac{1}{3} ln e^{3 x} + 1

\int \frac{2 e ^{x}}{e ^{3 x} + 1} d x = 2 (\frac{1}{3} ln ∣ e^{x} + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 e ^{2 x} - 4 e ^{x}}{3} + 23 arctan (\frac{2 e ^{x} - 1}{3})))

= \frac{1}{3} ln e^{3 x} + 1 + 2 (\frac{1}{3} ln ∣ e^{x} + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 e ^{2 x} - 4 e ^{x}}{3} + 23 arctan (\frac{2 e ^{x} - 1}{3})))

Add a constant to the solution

= \frac{1}{3} ln e^{3 x} + 1 + 2 (\frac{1}{3} ln ∣ e^{x} + 1 ∣ + \frac{1}{6} (- ln \frac{4}{3} + \frac{4 e ^{2 x} - 4 e ^{x}}{3} + 23 arctan (\frac{2 e ^{x} - 1}{3}))) + C

\frac{\partial}{\partial x} (5 y e^{4 x})

(- 2, 3), (2, - 2)

\frac{d v}{d t} = v^{2} + 5 v + 2

\frac{d}{d x} (\frac{4}{3} x^{3} + \frac{1}{2} x - 4)

\int \frac{6 e ^{6 t}}{9 + e ^{6 t}} d t

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