What is the integral of 2^{-3x} ?

The integral of 2^{-3x} is -(8^{-x})/(3ln(2))+C

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integral of 2^{-3x}

\int 2^{- 3 x} d x

- \frac{8 ^{- x}}{3 ln ( 2 )} + C

Solution steps

\int 2^{- 3 x} d x

Apply u-substitution

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

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

Take the constant out:

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

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

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

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

Substitute back

u = - 3 x

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

Simplify

- \frac{1}{3} \cdot \frac{2 ^{- 3 x}}{ln ( 2 )} : - \frac{8 ^{- x}}{3 ln ( 2 )}

= - \frac{8 ^{- x}}{3 ln ( 2 )}

Add a constant to the solution

= - \frac{8 ^{- x}}{3 ln ( 2 )} + C

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What is the integral of 2^{-3x} ?
The integral of 2^{-3x} is -(8^{-x})/(3ln(2))+C