What is the integral of 3^{(2x+4)} ?

The integral of 3^{(2x+4)} is (3^{2x+4})/(2ln(3))+C

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integral of 3^{(2x+4)}

\int 3^{(2 x + 4)} d x

\frac{3 ^{2 x + 4}}{2 ln ( 3 )} + C

Solution steps

\int 3^{2 x + 4} d x

Apply u-substitution

: \int 3^{u} \cdot \frac{1}{2} d u

= \int 3^{u} \cdot \frac{1}{2} d u

Take the constant out:

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

= \frac{1}{2} \cdot \int 3^{u} d u

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

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

Substitute back

u = 2 x + 4

= \frac{1}{2} \cdot \frac{3 ^{2 x + 4}}{ln ( 3 )}

Simplify

\frac{1}{2} \cdot \frac{3 ^{2 x + 4}}{ln ( 3 )} : \frac{3 ^{2 x + 4}}{2 ln ( 3 )}

= \frac{3 ^{2 x + 4}}{2 ln ( 3 )}

Add a constant to the solution

= \frac{3 ^{2 x + 4}}{2 ln ( 3 )} + C

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What is the integral of 3^{(2x+4)} ?
The integral of 3^{(2x+4)} is (3^{2x+4})/(2ln(3))+C