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

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

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integral of 2^xe^{3x}

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

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

Solution steps

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

Apply u-substitution

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

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

Take the constant out:

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

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

Apply the Power Rule

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

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

Substitute back

u = 2^{x}

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

Simplify

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

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

Add a constant to the solution

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

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