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

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

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

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

\frac{1}{3 ( 9 + 4 e )} e^{3 x} + C

Solution steps

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

Take the constant out:

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

= \frac{1}{9 + 4 e} \cdot \int e^{3 x} d x

Apply u-substitution

= \frac{1}{9 + 4 e} \cdot \int e^{u} \frac{1}{3} d u

Take the constant out:

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

= \frac{1}{9 + 4 e} \cdot \frac{1}{3} \cdot \int e^{u} d u

Use the common integral:

\int e^{u} d u = e^{u}

= \frac{1}{9 + 4 e} \cdot \frac{1}{3} e^{u}

Substitute back

u = 3 x

= \frac{1}{9 + 4 e} \cdot \frac{1}{3} e^{3 x}

Simplify

\frac{1}{9 + 4 e} \cdot \frac{1}{3} e^{3 x} : \frac{1}{3 ( 9 + 4 e )} e^{3 x}

= \frac{1}{3 ( 9 + 4 e )} e^{3 x}

Add a constant to the solution

= \frac{1}{3 ( 9 + 4 e )} e^{3 x} + C

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