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

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

\int \frac{( e ^{3 x} )}{1 + e ^{6 x}} d x

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

Solution steps

\int \frac{e ^{3 x}}{1 + e ^{6 x}} d x

Apply u-substitution

: \int \frac{1}{6 u u - 1} d u

= \int \frac{1}{6 u u - 1} d u

Take the constant out:

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

= \frac{1}{6} \cdot \int \frac{1}{u u - 1} d u

Apply u-substitution

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

= \frac{1}{6} \cdot \int \frac{2}{v ^{2} + 1} d v

Take the constant out:

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

= \frac{1}{6} \cdot 2 \cdot \int \frac{1}{v ^{2} + 1} d v

Use the common integral:

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

= \frac{1}{6} \cdot 2 arctan (v)

Substitute back

= \frac{1}{6} \cdot 2 arctan (1 + e^{6 x} - 1)

Simplify

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

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

Add a constant to the solution

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

\int \frac{- 2 x - 1}{x ^{2} + 1} d x

\int x (21 - x)^{\frac{7}{2}} d x

\int - \frac{8}{x ^{9}} d x

\int \frac{x ^{2} - 24 x}{( x - 12 ) ^{2}} d x

\int 3 e^{\frac{1}{4} x} d x

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