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

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

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integral of (e^{2x}+e^{5x})/(e^{3x)}

\int \frac{e ^{2 x} + e ^{5 x}}{e ^{3 x}} d x

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

Solution steps

\int \frac{e ^{2 x} + e ^{5 x}}{e ^{3 x}} d x

Simplify

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

= \int \frac{1 + e ^{3 x}}{e ^{x}} d x

Apply u-substitution

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

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

Expand

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

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

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

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

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

Substitute back

u = e^{x}

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

Simplify

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

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

Add a constant to the solution

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

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