What is the integral of (e^{3x}+5e^{2x}-4e^x)/(e^x) ?

The integral of (e^{3x}+5e^{2x}-4e^x)/(e^x) is 1/2 e^{2x}+5e^x-4x+C

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integral of (e^{3x}+5e^{2x}-4e^x)/(e^x)

\int \frac{e ^{3 x} + 5 e ^{2 x} - 4 e ^{x}}{e ^{x}} d x

\frac{1}{2} e^{2 x} + 5 e^{x} - 4 x + C

Solution steps

\int \frac{e ^{3 x} + 5 e ^{2 x} - 4 e ^{x}}{e ^{x}} d x

Expand

\frac{e ^{3 x} + 5 e ^{2 x} - 4 e ^{x}}{e ^{x}} : \frac{e ^{3 x}}{e ^{x}} + \frac{5 e ^{2 x}}{e ^{x}} - \frac{4 e ^{x}}{e ^{x}}

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{e ^{3 x}}{e ^{x}} d x + \int \frac{5 e ^{2 x}}{e ^{x}} d x - \int \frac{4 e ^{x}}{e ^{x}} d x

\int \frac{e ^{3 x}}{e ^{x}} d x = \frac{1}{2} e^{2 x}

\int \frac{5 e ^{2 x}}{e ^{x}} d x = 5 e^{x}

\int \frac{4 e ^{x}}{e ^{x}} d x = 4 x

= \frac{1}{2} e^{2 x} + 5 e^{x} - 4 x

Add a constant to the solution

= \frac{1}{2} e^{2 x} + 5 e^{x} - 4 x + C

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What is the integral of (e^{3x}+5e^{2x}-4e^x)/(e^x) ?
The integral of (e^{3x}+5e^{2x}-4e^x)/(e^x) is 1/2 e^{2x}+5e^x-4x+C