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

The integral of (e/3)^{3x} is (e^{3x})/((1-ln(3))*3^{1+3x)}+C

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integral of (e/3)^{3x}

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

\frac{e ^{3 x}}{( 1 - ln ( 3 ) ) \cdot 3 ^{1 + 3 x}} + C

Solution steps

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

Apply u-substitution

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

= \int (\frac{e}{3})^{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}{3} \cdot \int (\frac{e}{3})^{u} d u

\int a^{x} d x = \frac{a ^{x}}{ln a}

= \frac{1}{3} \cdot \frac{( \frac{e}{3} ) ^{u}}{ln ( \frac{e}{3} )}

Substitute back

u = 3 x

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

Simplify

\frac{1}{3} \cdot \frac{( \frac{e}{3} ) ^{3 x}}{ln ( \frac{e}{3} )} : \frac{e ^{3 x}}{( 1 - ln ( 3 ) ) \cdot 3 ^{1 + 3 x}}

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

Add a constant to the solution

= \frac{e ^{3 x}}{( 1 - ln ( 3 ) ) \cdot 3 ^{1 + 3 x}} + C

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What is the integral of (e/3)^{3x} ?
The integral of (e/3)^{3x} is (e^{3x})/((1-ln(3))*3^{1+3x)}+C