What is the integral of (2x^2-6x)e^{2x} ?

The integral of (2x^2-6x)e^{2x} is e^{2x}x^2-4e^{2x}x+2e^{2x}+C

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integral of (2x^2-6x)e^{2x}

\int (2 x^{2} - 6 x) e^{2 x} d x

e^{2 x} x^{2} - 4 e^{2 x} x + 2 e^{2 x} + C

Solution steps

\int (2 x^{2} - 6 x) e^{2 x} d x

Expand

(2 x^{2} - 6 x) e^{2 x} : 2 e^{2 x} x^{2} - 6 e^{2 x} x

= \int 2 e^{2 x} x^{2} - 6 e^{2 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 2 e^{2 x} x^{2} d x - \int 6 e^{2 x} x d x

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

\int 6 e^{2 x} x d x = 3 e^{2 x} x - \frac{3}{2} e^{2 x}

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

Simplify

e^{2 x} x^{2} - e^{2 x} x + \frac{1}{2} e^{2 x} - (3 e^{2 x} x - \frac{3}{2} e^{2 x}) : e^{2 x} x^{2} - 4 e^{2 x} x + 2 e^{2 x}

= e^{2 x} x^{2} - 4 e^{2 x} x + 2 e^{2 x}

Add a constant to the solution

= e^{2 x} x^{2} - 4 e^{2 x} x + 2 e^{2 x} + C

\frac{\partial}{\partial y} (\frac{x + y}{x ^{2} + y ^{2}})

\frac{d}{d x} (\frac{x ^{4}}{4} - \frac{x ^{3}}{3} - 3 x^{2} + 4)

\int (4 - 3 x)^{7} d x

cos (2 (t - \frac{π}{2}))

\int - 6 sin (x) d x

What is the integral of (2x^2-6x)e^{2x} ?
The integral of (2x^2-6x)e^{2x} is e^{2x}x^2-4e^{2x}x+2e^{2x}+C