What is the integral of e^{3x}-18e^{-4x} ?

The integral of e^{3x}-18e^{-4x} is 1/3 e^{3x}+9/2 e^{-4x}+C

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integral of e^{3x}-18e^{-4x}

\int e^{3 x} - 18 e^{- 4 x} d x

\frac{1}{3} e^{3 x} + \frac{9}{2} e^{- 4 x} + C

Solution steps

\int e^{3 x} - 18 e^{- 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^{3 x} d x - \int 18 e^{- 4 x} d x

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

\int 18 e^{- 4 x} d x = - \frac{9}{2} e^{- 4 x}

= \frac{1}{3} e^{3 x} - (- \frac{9}{2} e^{- 4 x})

Simplify

= \frac{1}{3} e^{3 x} + \frac{9}{2} e^{- 4 x}

Add a constant to the solution

= \frac{1}{3} e^{3 x} + \frac{9}{2} e^{- 4 x} + C

What is the integral of e^{3x}-18e^{-4x} ?
The integral of e^{3x}-18e^{-4x} is 1/3 e^{3x}+9/2 e^{-4x}+C