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

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

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integral of 3^{-x}

\int 3^{- x} d x

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

Solution steps

\int 3^{- x} d x

Apply u-substitution

: \int - 3^{u} d u

= \int - 3^{u} d u

Take the constant out:

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

= - \int 3^{u} d u

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

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

Substitute back

u = - x

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

Add a constant to the solution

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

x^{^{''}} = - x + 3 x^{^{'}}

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