What is the integral of (e^{6x}+e^{-9x}) ?

The integral of (e^{6x}+e^{-9x}) is 1/6 e^{6x}-1/9 e^{-9x}+C

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integral of (e^{6x}+e^{-9x})

\int (e^{6 x} + e^{- 9 x}) d x

\frac{1}{6} e^{6 x} - \frac{1}{9} e^{- 9 x} + C

Solution steps

\int e^{6 x} + e^{- 9 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^{6 x} d x + \int e^{- 9 x} d x

\int e^{6 x} d x = \frac{1}{6} e^{6 x}

\int e^{- 9 x} d x = - \frac{1}{9} e^{- 9 x}

= \frac{1}{6} e^{6 x} - \frac{1}{9} e^{- 9 x}

Add a constant to the solution

= \frac{1}{6} e^{6 x} - \frac{1}{9} e^{- 9 x} + C

\int x^{2} \cdot cos (n x) d x

\frac{d}{d x} (\frac{2 x}{x - 4})

\frac{d}{d x} (\frac{2 x}{x ^{2} + 9})

\frac{3 s - 15}{2 s ^{2} - 4 s + 10}

\frac{d}{d x} (\frac{x ^{2} + 1}{x ^{2} - 4 x + 3})

What is the integral of (e^{6x}+e^{-9x}) ?
The integral of (e^{6x}+e^{-9x}) is 1/6 e^{6x}-1/9 e^{-9x}+C