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

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

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

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

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

Solution steps

\int e^{2 x} + e^{- 3 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^{2 x} d x + \int e^{- 3 x} d x

\int e^{2 x} d x = \frac{1}{2} e^{2 x}

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

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

Add a constant to the solution

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

(ln (x))^{5}

\int \frac{x - 2 x ^{2} + 1}{x} d x

r \to \infty lim (ln (r))

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

f (x) = (x^{3} - 7^{x} + 3), at x = 1

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