What is the integral of (9^{2x+5})/3 ?

The integral of (9^{2x+5})/3 is (3^{4x+9})/(4ln(3))+C

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integral of (9^{2x+5})/3

\int \frac{9 ^{2 x + 5}}{3} d x

\frac{3 ^{4 x + 9}}{4 ln ( 3 )} + C

Solution steps

\int \frac{9 ^{2 x + 5}}{3} d x

Take the constant out:

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

= \frac{1}{3} \cdot \int 9^{2 x + 5} d x

Apply u-substitution

: \int 9^{u} \cdot \frac{1}{2} d u

= \frac{1}{3} \cdot \int 9^{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}{3} \cdot \frac{1}{2} \cdot \int 9^{u} d u

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

= \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{9 ^{u}}{ln ( 9 )}

Substitute back

u = 2 x + 5

= \frac{1}{3} \cdot \frac{1}{2} \cdot \frac{9 ^{2 x + 5}}{ln ( 9 )}

Simplify

\frac{1}{3} \cdot \frac{1}{2} \cdot \frac{9 ^{2 x + 5}}{ln ( 9 )} : \frac{3 ^{4 x + 9}}{4 ln ( 3 )}

= \frac{3 ^{4 x + 9}}{4 ln ( 3 )}

Add a constant to the solution

= \frac{3 ^{4 x + 9}}{4 ln ( 3 )} + C

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What is the integral of (9^{2x+5})/3 ?
The integral of (9^{2x+5})/3 is (3^{4x+9})/(4ln(3))+C