What is the integral of (5-e^x)/(e^{3x)} ?

The integral of (5-e^x)/(e^{3x)} is -5/(3e^{3x)}+1/(2e^{2x)}+C

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integral of (5-e^x)/(e^{3x)}

\int \frac{5 - e ^{x}}{e ^{3 x}} d x

- \frac{5}{3 e ^{3 x}} + \frac{1}{2 e ^{2 x}} + C

Solution steps

\int \frac{5 - e ^{x}}{e ^{3 x}} d x

Apply u-substitution

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

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

Expand

\frac{5 - u}{u ^{4}} : \frac{5}{u ^{4}} - \frac{1}{u ^{3}}

= \int \frac{5}{u ^{4}} - \frac{1}{u ^{3}} 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 ^{4}} d u - \int \frac{1}{u ^{3}} d u

\int \frac{5}{u ^{4}} d u = - \frac{5}{3 u ^{3}}

\int \frac{1}{u ^{3}} d u = - \frac{1}{2 u ^{2}}

= - \frac{5}{3 u ^{3}} - (- \frac{1}{2 u ^{2}})

Substitute back

u = e^{x}

= - \frac{5}{3 ( e ^{x} ) ^{3}} - (- \frac{1}{2 ( e ^{x} ) ^{2}})

Simplify

- \frac{5}{3 ( e ^{x} ) ^{3}} - (- \frac{1}{2 ( e ^{x} ) ^{2}}) : - \frac{5}{3 e ^{3 x}} + \frac{1}{2 e ^{2 x}}

= - \frac{5}{3 e ^{3 x}} + \frac{1}{2 e ^{2 x}}

Add a constant to the solution

= - \frac{5}{3 e ^{3 x}} + \frac{1}{2 e ^{2 x}} + C

\int sin (x) 2 d x

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

\int x^{3} (x + 1)^{5} d x

y = x^{2} - 9, y = 7

(\frac{x}{7} + \frac{7}{x})^{7}

What is the integral of (5-e^x)/(e^{3x)} ?
The integral of (5-e^x)/(e^{3x)} is -5/(3e^{3x)}+1/(2e^{2x)}+C