What is the integral of (2-e^x)/(6e^{4x)} ?

The integral of (2-e^x)/(6e^{4x)} is 1/6 (-1/2 e^{-4x}+1/3 e^{-3x})+C

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integral of (2-e^x)/(6e^{4x)}

\int \frac{2 - e ^{x}}{6 e ^{4 x}} d x

\frac{1}{6} (- \frac{1}{2} e^{- 4 x} + \frac{1}{3} e^{- 3 x}) + C

Solution steps

\int \frac{2 - e ^{x}}{6 e ^{4 x}} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= \frac{1}{6} \cdot \int \frac{2 - e ^{x}}{e ^{4 x}} d x

Expand

\frac{2 - e ^{x}}{e ^{4 x}} : \frac{2}{e ^{4 x}} - \frac{1}{e ^{3 x}}

= \frac{1}{6} \cdot \int \frac{2}{e ^{4 x}} - \frac{1}{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

= \frac{1}{6} (\int \frac{2}{e ^{4 x}} d x - \int \frac{1}{e ^{3 x}} d x)

\int \frac{2}{e ^{4 x}} d x = - \frac{1}{2} e^{- 4 x}

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

= \frac{1}{6} (- \frac{1}{2} e^{- 4 x} - (- \frac{1}{3} e^{- 3 x}))

Simplify

= \frac{1}{6} (- \frac{1}{2} e^{- 4 x} + \frac{1}{3} e^{- 3 x})

Add a constant to the solution

= \frac{1}{6} (- \frac{1}{2} e^{- 4 x} + \frac{1}{3} e^{- 3 x}) + C

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What is the integral of (2-e^x)/(6e^{4x)} ?
The integral of (2-e^x)/(6e^{4x)} is 1/6 (-1/2 e^{-4x}+1/3 e^{-3x})+C