What is the integral of (e^{3x})/(9+e^{6x)} ?

The integral of (e^{3x})/(9+e^{6x)} is 1/9 arctan(1/3 e^{3x})+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})/(9+e^{6x)}

\int \frac{e ^{3 x}}{9 + e ^{6 x}} d x

\frac{1}{9} arctan (\frac{1}{3} e^{3 x}) + C

Solution steps

\int \frac{e ^{3 x}}{9 + e ^{6 x}} d x

Apply u-substitution

: \int \frac{e ^{u}}{3 ( 9 + e ^{2 u} )} d u

= \int \frac{e ^{u}}{3 ( 9 + e ^{2 u} )} d u

Take the constant out:

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

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

Apply u-substitution

: \int \frac{1}{9 + v ^{2}} d v

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

Apply Integral Substitution

= \frac{1}{3} \cdot \int \frac{1}{3 ( w ^{2} + 1 )} d w

Take the constant out:

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

= \frac{1}{3} \cdot \frac{1}{3} \cdot \int \frac{1}{w ^{2} + 1} d w

Use the common integral:

\int \frac{1}{w ^{2} + 1} d w = arctan (w)

= \frac{1}{3} \cdot \frac{1}{3} arctan (w)

Substitute back

= \frac{1}{3} \cdot \frac{1}{3} arctan (\frac{e ^{3 x}}{3})

Simplify

\frac{1}{3} \cdot \frac{1}{3} arctan (\frac{e ^{3 x}}{3}) : \frac{1}{9} arctan (\frac{1}{3} e^{3 x})

= \frac{1}{9} arctan (\frac{1}{3} e^{3 x})

Add a constant to the solution

= \frac{1}{9} arctan (\frac{1}{3} e^{3 x}) + C

x \to 3 lim (1 0^{x^{2} - 4 x})

\int sin (t) + cos (t) d t

15 x^{2} - 4 y^{^{'}} = 0

\frac{\partial}{\partial x} (\frac{y}{h ^{2} x})

y^{^{'}} = 2 y (1 - 3 y)

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