What is the integral of (e^{-9x}+2^x) ?

The integral of (e^{-9x}+2^x) is -1/9 e^{-9x}+(2^x)/(ln(2))+C

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integral of (e^{-9x}+2^x)

\int (e^{- 9 x} + 2^{x}) d x

- \frac{1}{9} e^{- 9 x} + \frac{2 ^{x}}{ln ( 2 )} + C

Solution steps

\int e^{- 9 x} + 2^{x} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= \int e^{- 9 x} d x + \int 2^{x} d x

\int e^{- 9 x} d x = - \frac{1}{9} e^{- 9 x}

\int 2^{x} d x = \frac{2 ^{x}}{ln ( 2 )}

= - \frac{1}{9} e^{- 9 x} + \frac{2 ^{x}}{ln ( 2 )}

Add a constant to the solution

= - \frac{1}{9} e^{- 9 x} + \frac{2 ^{x}}{ln ( 2 )} + C

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What is the integral of (e^{-9x}+2^x) ?
The integral of (e^{-9x}+2^x) is -1/9 e^{-9x}+(2^x)/(ln(2))+C