What is the integral of 9^{-x} ?

The integral of 9^{-x} is -(9^{-x})/(2ln(3))+C

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integral of 9^{-x}

\int 9^{- x} d x

- \frac{9 ^{- x}}{2 ln ( 3 )} + C

Solution steps

\int 9^{- x} d x

Apply u-substitution

: \int - 9^{u} d u

= \int - 9^{u} d u

Take the constant out:

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

= - \int 9^{u} d u

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

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

Substitute back

u = - x

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

Simplify

ln (9) : 2 ln (3)

= - \frac{9 ^{- x}}{2 ln ( 3 )}

Add a constant to the solution

= - \frac{9 ^{- x}}{2 ln ( 3 )} + C

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