What is the integral of (e^{2x})/(1+6e^{2x)+9e^{4x}} ?

The integral of (e^{2x})/(1+6e^{2x)+9e^{4x}} is -1/(6(3e^{2x)+1)}+C

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral of (e^{2x})/(1+6e^{2x)+9e^{4x}}

\int \frac{e ^{2 x}}{1 + 6 e ^{2 x} + 9 e ^{4 x}} d x

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

Solution steps

\int \frac{e ^{2 x}}{1 + 6 e ^{2 x} + 9 e ^{4 x}} d x

Apply u-substitution

= \int \frac{1}{2 ( 1 + 6 u + 9 u ^{2} )} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int \frac{1}{1 + 6 u + 9 u ^{2}} d u

Complete the square

1 + 6 u + 9 u^{2} : 9 (u + \frac{1}{3})^{2}

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

Apply u-substitution

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

Take the constant out:

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

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

Apply the Power Rule

= \frac{1}{2} \cdot \frac{1}{9} (- \frac{1}{v})

Substitute back

= \frac{1}{2} \cdot \frac{1}{9} (- \frac{1}{e ^{2 x} + \frac{1}{3}})

Simplify

\frac{1}{2} \cdot \frac{1}{9} (- \frac{1}{e ^{2 x} + \frac{1}{3}}) : - \frac{1}{6 ( 3 e ^{2 x} + 1 )}

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

Add a constant to the solution

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

What is the integral of (e^{2x})/(1+6e^{2x)+9e^{4x}} ?
The integral of (e^{2x})/(1+6e^{2x)+9e^{4x}} is -1/(6(3e^{2x)+1)}+C