What is the integral of (6e^{3x}-8e^{-2x}) ?

The integral of (6e^{3x}-8e^{-2x}) is 2e^{3x}+4e^{-2x}+C

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integral of (6e^{3x}-8e^{-2x})

\int (6 e^{3 x} - 8 e^{- 2 x}) d x

2 e^{3 x} + 4 e^{- 2 x} + C

Solution steps

\int 6 e^{3 x} - 8 e^{- 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 6 e^{3 x} d x - \int 8 e^{- 2 x} d x

\int 6 e^{3 x} d x = 2 e^{3 x}

\int 8 e^{- 2 x} d x = - 4 e^{- 2 x}

= 2 e^{3 x} - (- 4 e^{- 2 x})

Simplify

= 2 e^{3 x} + 4 e^{- 2 x}

Add a constant to the solution

= 2 e^{3 x} + 4 e^{- 2 x} + C

What is the integral of (6e^{3x}-8e^{-2x}) ?
The integral of (6e^{3x}-8e^{-2x}) is 2e^{3x}+4e^{-2x}+C