What is the integral of (9-e^x)/(e^{5x)} ?

The integral of (9-e^x)/(e^{5x)} is -9/(5e^{5x)}+1/(4e^{4x)}+C

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integral of (9-e^x)/(e^{5x)}

\int \frac{9 - e ^{x}}{e ^{5 x}} d x

- \frac{9}{5 e ^{5 x}} + \frac{1}{4 e ^{4 x}} + C

Solution steps

\int \frac{9 - e ^{x}}{e ^{5 x}} d x

Apply u-substitution

: \int \frac{9 - u}{u ^{6}} d u

= \int \frac{9 - u}{u ^{6}} d u

Expand

\frac{9 - u}{u ^{6}} : \frac{9}{u ^{6}} - \frac{1}{u ^{5}}

= \int \frac{9}{u ^{6}} - \frac{1}{u ^{5}} d u

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{9}{u ^{6}} d u - \int \frac{1}{u ^{5}} d u

\int \frac{9}{u ^{6}} d u = - \frac{9}{5 u ^{5}}

\int \frac{1}{u ^{5}} d u = - \frac{1}{4 u ^{4}}

= - \frac{9}{5 u ^{5}} - (- \frac{1}{4 u ^{4}})

Substitute back

u = e^{x}

= - \frac{9}{5 ( e ^{x} ) ^{5}} - (- \frac{1}{4 ( e ^{x} ) ^{4}})

Simplify

- \frac{9}{5 ( e ^{x} ) ^{5}} - (- \frac{1}{4 ( e ^{x} ) ^{4}}) : - \frac{9}{5 e ^{5 x}} + \frac{1}{4 e ^{4 x}}

= - \frac{9}{5 e ^{5 x}} + \frac{1}{4 e ^{4 x}}

Add a constant to the solution

= - \frac{9}{5 e ^{5 x}} + \frac{1}{4 e ^{4 x}} + C

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What is the integral of (9-e^x)/(e^{5x)} ?
The integral of (9-e^x)/(e^{5x)} is -9/(5e^{5x)}+1/(4e^{4x)}+C