What is the integral of 9^xe^{2x} ?

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

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integral of 9^xe^{2x}

\int 9^{x} e^{2 x} d x

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

Solution steps

\int 9^{x} e^{2 x} d x

Apply u-substitution

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

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

Apply the Power Rule

: \frac{1}{1 + ln ( 3 )} ln (3) u^{\frac{1 + l n ( 3 )}{l n ( 3 )}}

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

Substitute back

u = 9^{x}

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

Simplify

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

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

Add a constant to the solution

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

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