What is the integral of 1/9 e^{3x} ?

The integral of 1/9 e^{3x} is 1/27 e^{3x}+C

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integral of 1/9 e^{3x}

\int \frac{1}{9} e^{3 x} d x

\frac{1}{27} e^{3 x} + C

Solution steps

\int \frac{1}{9} e^{3 x} d x

Take the constant out:

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

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

Apply u-substitution

: \int e^{u} \frac{1}{3} d u

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

Use the common integral:

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

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

Substitute back

u = 3 x

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

Simplify

\frac{1}{9} \cdot \frac{1}{3} e^{3 x} : \frac{1}{27} e^{3 x}

= \frac{1}{27} e^{3 x}

Add a constant to the solution

= \frac{1}{27} e^{3 x} + C

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What is the integral of 1/9 e^{3x} ?
The integral of 1/9 e^{3x} is 1/27 e^{3x}+C