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

The integral of (1-e^{-x})/(e^{3x)} is -1/3 e^{-3x}+1/4 e^{-4x}+C

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integral of (1-e^{-x})/(e^{3x)}

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

- \frac{1}{3} e^{- 3 x} + \frac{1}{4} e^{- 4 x} + C

Solution steps

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

Expand

\frac{1 - e ^{- x}}{e ^{3 x}} : \frac{1}{e ^{3 x}} - \frac{1}{e ^{4 x}}

= \int \frac{1}{e ^{3 x}} - \frac{1}{e ^{4 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{1}{e ^{3 x}} d x - \int \frac{1}{e ^{4 x}} d x

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

\int \frac{1}{e ^{4 x}} d x = - \frac{1}{4} e^{- 4 x}

= - \frac{1}{3} e^{- 3 x} - (- \frac{1}{4} e^{- 4 x})

Simplify

= - \frac{1}{3} e^{- 3 x} + \frac{1}{4} e^{- 4 x}

Add a constant to the solution

= - \frac{1}{3} e^{- 3 x} + \frac{1}{4} e^{- 4 x} + C

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What is the integral of (1-e^{-x})/(e^{3x)} ?
The integral of (1-e^{-x})/(e^{3x)} is -1/3 e^{-3x}+1/4 e^{-4x}+C