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

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

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integral of a^{(3x+3)}

\int a^{(3 x + 3)} d x

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

Solution steps

\int a^{3 x + 3} d x

a^{3 x + 3} = a^{3} a^{3 x}

= \int a^{3} a^{3 x} d x

Take the constant out:

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

= a^{3} \cdot \int a^{3 x} d x

Apply u-substitution

: \int a^{u} \frac{1}{3} d u

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

Take the constant out:

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

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

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

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

Substitute back

u = 3 x

= a^{3} \frac{1}{3} \cdot \frac{a ^{3 x}}{ln ( a )}

Simplify

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

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

Add a constant to the solution

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

y^{'} = y (3 - y) - \frac{9}{4}

y = \frac{( 3 x ^{2} + 4 ) ( 2 x + 1 )}{5 x - 9}

f (z) = \frac{z - 8}{z + 8}

x \to 3 lim (\frac{5}{x ^{2} - 9})

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