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

The integral of 3^{-6x} is -(3^{-6x-1})/(2ln(3))+C

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integral of 3^{-6x}

\int 3^{- 6 x} d x

- \frac{3 ^{- 6 x - 1}}{2 ln ( 3 )} + C

Solution steps

\int 3^{- 6 x} d x

Simplify

3^{- 6 x} : 72 9^{- x}

= \int 72 9^{- x} d x

Apply u-substitution

: \int - 72 9^{u} d u

= \int - 72 9^{u} d u

Take the constant out:

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

= - \int 72 9^{u} d u

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

= - \frac{72 9 ^{u}}{ln ( 729 )}

Substitute back

u = - x

= - \frac{72 9 ^{- x}}{ln ( 729 )}

Simplify

- \frac{72 9 ^{- x}}{ln ( 729 )} : - \frac{3 ^{- 6 x - 1}}{2 ln ( 3 )}

= - \frac{3 ^{- 6 x - 1}}{2 ln ( 3 )}

Add a constant to the solution

= - \frac{3 ^{- 6 x - 1}}{2 ln ( 3 )} + C

\frac{3}{s ( s ^{2} + 2 s + 4 )}

\int y^{- 5} d y

y^{^{'}} = y^{2} cos (x)

x \to - 1 lim ((x + 2 x^{5})^{3})

\int (e^{x} - e^{5}) (e^{x} + e^{- 3}) d x

What is the integral of 3^{-6x} ?
The integral of 3^{-6x} is -(3^{-6x-1})/(2ln(3))+C