What is the integral of (e^{3x})/(9+4e^{6x)} ?

The integral of (e^{3x})/(9+4e^{6x)} is 1/18 arctan((2e^{3x})/3)+C

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integral of (e^{3x})/(9+4e^{6x)}

\int \frac{e ^{3 x}}{9 + 4 e ^{6 x}} d x

\frac{1}{18} arctan (\frac{2 e ^{3 x}}{3}) + C

Solution steps

\int \frac{e ^{3 x}}{9 + 4 e ^{6 x}} d x

Apply u-substitution

: \int \frac{1}{12 u u - 9} d u

= \int \frac{1}{12 u u - 9} d u

Take the constant out:

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

= \frac{1}{12} \cdot \int \frac{1}{u u - 9} d u

Apply u-substitution

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

= \frac{1}{12} \cdot \int \frac{2}{v ^{2} + 9} d v

Take the constant out:

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

= \frac{1}{12} \cdot 2 \cdot \int \frac{1}{v ^{2} + 9} d v

Apply Integral Substitution

= \frac{1}{12} \cdot 2 \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}{12} \cdot 2 \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}{12} \cdot 2 \cdot \frac{1}{3} arctan (w)

Substitute back

= \frac{1}{12} \cdot 2 \cdot \frac{1}{3} arctan (\frac{9 + 4 e ^{6 x} - 9}{3})

Simplify

\frac{1}{12} \cdot 2 \cdot \frac{1}{3} arctan (\frac{9 + 4 e ^{6 x} - 9}{3}) : \frac{1}{18} arctan (\frac{2 e ^{3 x}}{3})

= \frac{1}{18} arctan (\frac{2 e ^{3 x}}{3})

Add a constant to the solution

= \frac{1}{18} arctan (\frac{2 e ^{3 x}}{3}) + C

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What is the integral of (e^{3x})/(9+4e^{6x)} ?
The integral of (e^{3x})/(9+4e^{6x)} is 1/18 arctan((2e^{3x})/3)+C