What is the integral of (e^{-4x}+7^x) ?

The integral of (e^{-4x}+7^x) is -1/4 e^{-4x}+(7^x)/(ln(7))+C

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integral of (e^{-4x}+7^x)

\int (e^{- 4 x} + 7^{x}) d x

- \frac{1}{4} e^{- 4 x} + \frac{7 ^{x}}{ln ( 7 )} + C

Solution steps

\int e^{- 4 x} + 7^{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^{- 4 x} d x + \int 7^{x} d x

\int e^{- 4 x} d x = - \frac{1}{4} e^{- 4 x}

\int 7^{x} d x = \frac{7 ^{x}}{ln ( 7 )}

= - \frac{1}{4} e^{- 4 x} + \frac{7 ^{x}}{ln ( 7 )}

Add a constant to the solution

= - \frac{1}{4} e^{- 4 x} + \frac{7 ^{x}}{ln ( 7 )} + C

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What is the integral of (e^{-4x}+7^x) ?
The integral of (e^{-4x}+7^x) is -1/4 e^{-4x}+(7^x)/(ln(7))+C