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

The integral of (e^{4x})/(1+e^x) is ((1+e^x)^3)/3-(3(1+e^x)^2)/2+3(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^{4x})/(1+e^x)

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

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

Solution steps

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

Apply u-substitution

: \int \frac{( u - 1 ) ^{3}}{u} d u

= \int \frac{( u - 1 ) ^{3}}{u} d u

Expand

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

= \int u^{2} - 3 u + 3 - \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^{2} d u - \int 3 u d u + \int 3 d u - \int \frac{1}{u} d u

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

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

\int 3 d u = 3 u

\int \frac{1}{u} d u = ln ∣ u ∣

= \frac{u ^{3}}{3} - \frac{3 u ^{2}}{2} + 3 u - ln ∣ u ∣

Substitute back

u = 1 + e^{x}

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

Add a constant to the solution

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

\int t (1 - ln^{2} (t)) d t

\int (- 2 sin (x) + 4 cos (x)) d x

\int x e^{- (x + y)} d x

\frac{d}{d x} ((7 - 3 x) (x + 5))

y^{^{'}} + 3 y = cos (x)

What is the integral of (e^{4x})/(1+e^x) ?
The integral of (e^{4x})/(1+e^x) is ((1+e^x)^3)/3-(3(1+e^x)^2)/2+3(1+e^x)-ln|1+e^x|+C