What is the integral of 3^{4x+6} ?

The integral of 3^{4x+6} is (3^{4x+6})/(4ln(3))+C

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integral of 3^{4x+6}

\int 3^{4 x + 6} d x

\frac{3 ^{4 x + 6}}{4 ln ( 3 )} + C

Solution steps

\int 3^{4 x + 6} d x

Apply u-substitution

: \int 3^{u} \cdot \frac{1}{4} d u

= \int 3^{u} \cdot \frac{1}{4} d u

Take the constant out:

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

= \frac{1}{4} \cdot \int 3^{u} d u

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

= \frac{1}{4} \cdot \frac{3 ^{u}}{ln ( 3 )}

Substitute back

u = 4 x + 6

= \frac{1}{4} \cdot \frac{3 ^{4 x + 6}}{ln ( 3 )}

Simplify

\frac{1}{4} \cdot \frac{3 ^{4 x + 6}}{ln ( 3 )} : \frac{3 ^{4 x + 6}}{4 ln ( 3 )}

= \frac{3 ^{4 x + 6}}{4 ln ( 3 )}

Add a constant to the solution

= \frac{3 ^{4 x + 6}}{4 ln ( 3 )} + C

\frac{\partial}{\partial x} (2 y + x^{2} + 1)

2 cos^{2} (t)

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

\frac{d}{d x} (x^{3} 1 - x^{2})

What is the integral of 3^{4x+6} ?
The integral of 3^{4x+6} is (3^{4x+6})/(4ln(3))+C