What is the integral of (2e^{9x})/(e^{3x)+1} ?

The integral of (2e^{9x})/(e^{3x)+1} is 1/3 (e^{6x}+2ln|e^{3x}+1|-2e^{3x}-3)+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

Popular >

integral of (2e^{9x})/(e^{3x)+1}

\int \frac{2 e ^{9 x}}{e ^{3 x} + 1} d x

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

Solution steps

\int \frac{2 e ^{9 x}}{e ^{3 x} + 1} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \int \frac{e ^{9 x}}{e ^{3 x} + 1} d x

Apply u-substitution

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

= 2 \cdot \int \frac{( u - 1 ) ^{2}}{3 u} d u

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 2 \cdot \frac{1}{3} \cdot \int \frac{( u - 1 ) ^{2}}{u} d u

Expand

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

= 2 \cdot \frac{1}{3} \cdot \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

= 2 \cdot \frac{1}{3} (\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 ∣

= 2 \cdot \frac{1}{3} (\frac{u ^{2}}{2} - 2 u + ln ∣ u ∣)

Substitute back

u = e^{3 x} + 1

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

Simplify

2 \cdot \frac{1}{3} (\frac{( e ^{3 x} + 1 ) ^{2}}{2} - 2 (e^{3 x} + 1) + ln e^{3 x} + 1) : \frac{1}{3} (e^{6 x} + 2 ln e^{3 x} + 1 - 2 e^{3 x} - 3)

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

Add a constant to the solution

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

\int (\frac{cos ( x )}{sin ( x )}) d x

\int (8 - x) (2 x + 8) d x

\frac{\partial}{\partial x} (\frac{1}{4} (x + y)^{4})

- 4 x + \frac{2 x d y}{d x} = \frac{2 d y}{d x} + 3 x^{2} + 2

y = \frac{1}{( 1 + x ^{2} ) , at ( 1 , \frac{1}{2} )}

What is the integral of (2e^{9x})/(e^{3x)+1} ?
The integral of (2e^{9x})/(e^{3x)+1} is 1/3 (e^{6x}+2ln|e^{3x}+1|-2e^{3x}-3)+C