What is the integral of (e^{-x}+4^x) ?

The integral of (e^{-x}+4^x) is -e^{-x}+(2^{2x-1})/(ln(2))+C

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integral of (e^{-x}+4^x)

\int (e^{- x} + 4^{x}) d x

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

Solution steps

\int e^{- x} + 4^{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^{- x} d x + \int 4^{x} d x

\int e^{- x} d x = - e^{- x}

\int 4^{x} d x = \frac{2 ^{2 x - 1}}{ln ( 2 )}

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

Add a constant to the solution

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

What is the integral of (e^{-x}+4^x) ?
The integral of (e^{-x}+4^x) is -e^{-x}+(2^{2x-1})/(ln(2))+C