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

\frac{d}{d x} (in (x))

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

2 x^{3} - x^{2}, 0, 1

\frac{d y}{d t} - 2 y = y^{4}

\frac{d}{d x} ((x - 3) (x^{2} + 6))

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