What is the integral of (5-e^x)/(e^{4x)} ?

The integral of (5-e^x)/(e^{4x)} is -5/(4e^{4x)}+1/(3e^{3x)}+C

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integral of (5-e^x)/(e^{4x)}

\int \frac{5 - e ^{x}}{e ^{4 x}} d x

- \frac{5}{4 e ^{4 x}} + \frac{1}{3 e ^{3 x}} + C

Solution steps

\int \frac{5 - e ^{x}}{e ^{4 x}} d x

Apply u-substitution

: \int \frac{5 - u}{u ^{5}} d u

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

Expand

\frac{5 - u}{u ^{5}} : \frac{5}{u ^{5}} - \frac{1}{u ^{4}}

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

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

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

= - \frac{5}{4 u ^{4}} - (- \frac{1}{3 u ^{3}})

Substitute back

u = e^{x}

= - \frac{5}{4 ( e ^{x} ) ^{4}} - (- \frac{1}{3 ( e ^{x} ) ^{3}})

Simplify

- \frac{5}{4 ( e ^{x} ) ^{4}} - (- \frac{1}{3 ( e ^{x} ) ^{3}}) : - \frac{5}{4 e ^{4 x}} + \frac{1}{3 e ^{3 x}}

= - \frac{5}{4 e ^{4 x}} + \frac{1}{3 e ^{3 x}}

Add a constant to the solution

= - \frac{5}{4 e ^{4 x}} + \frac{1}{3 e ^{3 x}} + C

\frac{\partial}{\partial x} (ln (x + 9 y + 5 z))

\frac{d}{d x} (4 a e^{4 x})

\int_{- \infty}^{2} x^{2} e^{x} d x

\int \frac{2}{4 x + 1} d x

y = \frac{x}{x + 3}, (4, \frac{2}{7})

What is the integral of (5-e^x)/(e^{4x)} ?
The integral of (5-e^x)/(e^{4x)} is -5/(4e^{4x)}+1/(3e^{3x)}+C