What is the integral of (e^{6x}+4)(e^{-2x}+18) ?

The integral of (e^{6x}+4)(e^{-2x}+18) is 1/4 e^{4x}+3e^{6x}-2e^{-2x}+72x+C

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integral of (e^{6x}+4)(e^{-2x}+18)

\int (e^{6 x} + 4) (e^{- 2 x} + 18) d x

\frac{1}{4} e^{4 x} + 3 e^{6 x} - 2 e^{- 2 x} + 72 x + C

Solution steps

\int (e^{6 x} + 4) (e^{- 2 x} + 18) d x

Expand

(e^{6 x} + 4) (e^{- 2 x} + 18) : e^{4 x} + 18 e^{6 x} + 4 e^{- 2 x} + 72

= \int e^{4 x} + 18 e^{6 x} + 4 e^{- 2 x} + 72 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 18 e^{6 x} d x + \int 4 e^{- 2 x} d x + \int 72 d x

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

\int 18 e^{6 x} d x = 3 e^{6 x}

\int 4 e^{- 2 x} d x = - 2 e^{- 2 x}

\int 72 d x = 72 x

= \frac{1}{4} e^{4 x} + 3 e^{6 x} - 2 e^{- 2 x} + 72 x

Add a constant to the solution

= \frac{1}{4} e^{4 x} + 3 e^{6 x} - 2 e^{- 2 x} + 72 x + C

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What is the integral of (e^{6x}+4)(e^{-2x}+18) ?
The integral of (e^{6x}+4)(e^{-2x}+18) is 1/4 e^{4x}+3e^{6x}-2e^{-2x}+72x+C