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

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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

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

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

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

Solution steps

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

Apply u-substitution

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

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

Expand

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

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

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

\int 2 d u = 2 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{2}}{2} - 2 u + ln ∣ u ∣

Substitute back

u = 1 + e^{x}

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

Add a constant to the solution

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

x \to 3 lim (x^{2} + 3 x - 1)

\int e^{2} cos (3 x) d x

\frac{d}{d x} (cos (arcsin (3 x)))

\frac{d x}{d t} = 6 - \frac{7 x}{600 + 2 t}

\int \frac{3 x - 5}{x ^{2}} d x

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