What is the integral of (2-e^x)/(e^{9x)} ?

The integral of (2-e^x)/(e^{9x)} is -2/9 e^{-9x}+1/8 e^{-8x}+C

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integral of (2-e^x)/(e^{9x)}

\int \frac{2 - e ^{x}}{e ^{9 x}} d x

- \frac{2}{9} e^{- 9 x} + \frac{1}{8} e^{- 8 x} + C

Solution steps

\int \frac{2 - e ^{x}}{e ^{9 x}} d x

Expand

\frac{2 - e ^{x}}{e ^{9 x}} : \frac{2}{e ^{9 x}} - \frac{1}{e ^{8 x}}

= \int \frac{2}{e ^{9 x}} - \frac{1}{e ^{8 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{2}{e ^{9 x}} d x - \int \frac{1}{e ^{8 x}} d x

\int \frac{2}{e ^{9 x}} d x = - \frac{2}{9} e^{- 9 x}

\int \frac{1}{e ^{8 x}} d x = - \frac{1}{8} e^{- 8 x}

= - \frac{2}{9} e^{- 9 x} - (- \frac{1}{8} e^{- 8 x})

Simplify

= - \frac{2}{9} e^{- 9 x} + \frac{1}{8} e^{- 8 x}

Add a constant to the solution

= - \frac{2}{9} e^{- 9 x} + \frac{1}{8} e^{- 8 x} + C

\int_{0}^{\frac{π}{2}} sec^{4} (\frac{t}{2}) d t

x \to 3 lim (\frac{x - 3}{x ^{2} - 9})

f (x) = (\frac{x + 7}{x + 9})^{6}

\int \frac{1}{4 x 16 x ^{2} - 25} d x

What is the integral of (2-e^x)/(e^{9x)} ?
The integral of (2-e^x)/(e^{9x)} is -2/9 e^{-9x}+1/8 e^{-8x}+C