What is the integral of (e^x+e^{2x}+e^{3x})/(e^{4x)} ?

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

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integral of (e^x+e^{2x}+e^{3x})/(e^{4x)}

\int \frac{e ^{x} + e ^{2 x} + e ^{3 x}}{e ^{4 x}} d x

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

Solution steps

\int \frac{e ^{x} + e ^{2 x} + e ^{3 x}}{e ^{4 x}} d x

Apply u-substitution

: \int \frac{1 + u + u ^{2}}{u ^{4}} d u

= \int \frac{1 + u + u ^{2}}{u ^{4}} d u

Expand

\frac{1 + u + u ^{2}}{u ^{4}} : \frac{1}{u ^{4}} + \frac{1}{u ^{3}} + \frac{1}{u ^{2}}

= \int \frac{1}{u ^{4}} + \frac{1}{u ^{3}} + \frac{1}{u ^{2}} 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{1}{u ^{4}} d u + \int \frac{1}{u ^{3}} d u + \int \frac{1}{u ^{2}} d u

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

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

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

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

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

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