What is the integral of (e^{7x}+12)(e^{-4x}+14) ?

The integral of (e^{7x}+12)(e^{-4x}+14) is 1/3 e^{3x}+2e^{7x}-3e^{-4x}+168x+C

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integral of (e^{7x}+12)(e^{-4x}+14)

\int (e^{7 x} + 12) (e^{- 4 x} + 14) d x

\frac{1}{3} e^{3 x} + 2 e^{7 x} - 3 e^{- 4 x} + 168 x + C

Solution steps

\int (e^{7 x} + 12) (e^{- 4 x} + 14) d x

Expand

(e^{7 x} + 12) (e^{- 4 x} + 14) : e^{3 x} + 14 e^{7 x} + 12 e^{- 4 x} + 168

= \int e^{3 x} + 14 e^{7 x} + 12 e^{- 4 x} + 168 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^{3 x} d x + \int 14 e^{7 x} d x + \int 12 e^{- 4 x} d x + \int 168 d x

\int e^{3 x} d x = \frac{1}{3} e^{3 x}

\int 14 e^{7 x} d x = 2 e^{7 x}

\int 12 e^{- 4 x} d x = - 3 e^{- 4 x}

\int 168 d x = 168 x

= \frac{1}{3} e^{3 x} + 2 e^{7 x} - 3 e^{- 4 x} + 168 x

Add a constant to the solution

= \frac{1}{3} e^{3 x} + 2 e^{7 x} - 3 e^{- 4 x} + 168 x + C

What is the integral of (e^{7x}+12)(e^{-4x}+14) ?
The integral of (e^{7x}+12)(e^{-4x}+14) is 1/3 e^{3x}+2e^{7x}-3e^{-4x}+168x+C