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

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

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

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

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

Solution steps

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

Expand

\frac{3 e ^{- x} + 2 e ^{3 x}}{e ^{2 x}} : \frac{3}{e ^{3 x}} + 2 e^{x}

= \int \frac{3}{e ^{3 x}} + 2 e^{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 \frac{3}{e ^{3 x}} d x + \int 2 e^{x} d x

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

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

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

Add a constant to the solution

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

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