What is the integral of (e^{2x}-2e^x+1)e^x ?

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

integral of (e^{2x}-2e^x+1)e^x

\int (e^{2 x} - 2 e^{x} + 1) e^{x} d x

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

Solution steps

\int (e^{2 x} - 2 e^{x} + 1) e^{x} d x

Expand

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

= \int e^{3 x} - 2 e^{2 x} + e^{x} d 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 e^{3 x} d x - \int 2 e^{2 x} d x + \int e^{x} d x

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

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

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

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

Add a constant to the solution

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

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