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

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

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

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

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

Solution steps

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

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

\int \frac{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

y^{^{''}} - 4 y^{^{'}} + 4 y = (x + 1) e^{2 x}

f (x) = \frac{1}{8 x}, at (x = 9)

n = 0 \sum \infty \frac{2 ^{n}}{( 2 n ) !} x^{2 n}

x \to 7 - lim (\frac{x - 7}{∣ x - 7 ∣})

\int_{0}^{9} \frac{11 e ^{x}}{x} d x

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