What is the integral of (x^2-8x+16)e^{2x} ?

The integral of (x^2-8x+16)e^{2x} is 1/2 e^{2x}x^2-9/2 e^{2x}x+41/4 e^{2x}+C

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integral of (x^2-8x+16)e^{2x}

\int (x^{2} - 8 x + 16) e^{2 x} d x

\frac{1}{2} e^{2 x} x^{2} - \frac{9}{2} e^{2 x} x + \frac{41}{4} e^{2 x} + C

Solution steps

\int (x^{2} - 8 x + 16) e^{2 x} d x

Expand

(x^{2} - 8 x + 16) e^{2 x} : e^{2 x} x^{2} - 8 e^{2 x} x + 16 e^{2 x}

= \int e^{2 x} x^{2} - 8 e^{2 x} x + 16 e^{2 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^{2 x} x^{2} d x - \int 8 e^{2 x} x d x + \int 16 e^{2 x} d x

\int e^{2 x} x^{2} d x = \frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})

\int 8 e^{2 x} x d x = 2 (2 e^{2 x} x - e^{2 x})

\int 16 e^{2 x} d x = 8 e^{2 x}

= \frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) - 2 (2 e^{2 x} x - e^{2 x}) + 8 e^{2 x}

Simplify

\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}) - 2 (2 e^{2 x} x - e^{2 x}) + 8 e^{2 x} : \frac{1}{2} e^{2 x} x^{2} - \frac{9}{2} e^{2 x} x + \frac{41}{4} e^{2 x}

= \frac{1}{2} e^{2 x} x^{2} - \frac{9}{2} e^{2 x} x + \frac{41}{4} e^{2 x}

Add a constant to the solution

= \frac{1}{2} e^{2 x} x^{2} - \frac{9}{2} e^{2 x} x + \frac{41}{4} e^{2 x} + C

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What is the integral of (x^2-8x+16)e^{2x} ?
The integral of (x^2-8x+16)e^{2x} is 1/2 e^{2x}x^2-9/2 e^{2x}x+41/4 e^{2x}+C