What is the integral of (x^3-3x)e^{6x} ?

The integral of (x^3-3x)e^{6x} is 1/6 e^{6x}x^3-1/12 e^{6x}x^2-17/36 e^{6x}x+17/216 e^{6x}+C

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integral of (x^3-3x)e^{6x}

\int (x^{3} - 3 x) e^{6 x} d x

\frac{1}{6} e^{6 x} x^{3} - \frac{1}{12} e^{6 x} x^{2} - \frac{17}{36} e^{6 x} x + \frac{17}{216} e^{6 x} + C

Solution steps

\int (x^{3} - 3 x) e^{6 x} d x

Expand

(x^{3} - 3 x) e^{6 x} : e^{6 x} x^{3} - 3 e^{6 x} x

= \int e^{6 x} x^{3} - 3 e^{6 x} 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^{6 x} x^{3} d x - \int 3 e^{6 x} x d x

\int e^{6 x} x^{3} d x = \frac{1}{216} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1)

\int 3 e^{6 x} x d x = \frac{1}{12} (6 e^{6 x} x - e^{6 x})

= \frac{1}{216} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1) - \frac{1}{12} (6 e^{6 x} x - e^{6 x})

Simplify

\frac{1}{216} e^{6 x} (36 x^{3} - 18 x^{2} + 6 x - 1) - \frac{1}{12} (6 e^{6 x} x - e^{6 x}) : \frac{1}{6} e^{6 x} x^{3} - \frac{1}{12} e^{6 x} x^{2} - \frac{17}{36} e^{6 x} x + \frac{17}{216} e^{6 x}

= \frac{1}{6} e^{6 x} x^{3} - \frac{1}{12} e^{6 x} x^{2} - \frac{17}{36} e^{6 x} x + \frac{17}{216} e^{6 x}

Add a constant to the solution

= \frac{1}{6} e^{6 x} x^{3} - \frac{1}{12} e^{6 x} x^{2} - \frac{17}{36} e^{6 x} x + \frac{17}{216} e^{6 x} + C

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What is the integral of (x^3-3x)e^{6x} ?
The integral of (x^3-3x)e^{6x} is 1/6 e^{6x}x^3-1/12 e^{6x}x^2-17/36 e^{6x}x+17/216 e^{6x}+C