What is the integral of 3^{2x}e^{3x} ?

The integral of 3^{2x}e^{3x} is 1/(2ln(3))*2/(3+2ln(3))ln(3)(9^x)^{(3+2ln(3))/(2ln(3))}+C

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integral of 3^{2x}e^{3x}

\int 3^{2 x} e^{3 x} d x

\frac{1}{2 ln ( 3 )} \cdot \frac{2}{3 + 2 ln ( 3 )} ln (3) (9^{x})^{\frac{3 + 2 l n ( 3 )}{2 l n ( 3 )}} + C

Solution steps

\int 3^{2 x} e^{3 x} d x

3^{2 x} = 9^{x}

= \int 9^{x} e^{3 x} d x

Apply u-substitution

: \int \frac{u ^{\frac{3}{2 l n ( 3 )}}}{2 ln ( 3 )} d u

= \int \frac{u ^{\frac{3}{2 l n ( 3 )}}}{2 ln ( 3 )} d u

Take the constant out:

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

= \frac{1}{2 ln ( 3 )} \cdot \int u^{\frac{3}{2 l n ( 3 )}} d u

Apply the Power Rule

: \frac{2}{3 + 2 ln ( 3 )} ln (3) u^{\frac{3 + 2 l n ( 3 )}{2 l n ( 3 )}}

= \frac{1}{2 ln ( 3 )} \cdot \frac{2}{3 + 2 ln ( 3 )} ln (3) u^{\frac{3 + 2 l n ( 3 )}{2 l n ( 3 )}}

Substitute back

u = 9^{x}

= \frac{1}{2 ln ( 3 )} \cdot \frac{2}{3 + 2 ln ( 3 )} ln (3) (9^{x})^{\frac{3 + 2 l n ( 3 )}{2 l n ( 3 )}}

Add a constant to the solution

= \frac{1}{2 ln ( 3 )} \cdot \frac{2}{3 + 2 ln ( 3 )} ln (3) (9^{x})^{\frac{3 + 2 l n ( 3 )}{2 l n ( 3 )}} + C

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