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

The integral of (3x^2-4x)e^{2x} is 3/2 e^{2x}x^2-7/2 e^{2x}x+7/4 e^{2x}+C

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

\int (3 x^{2} - 4 x) e^{2 x} d x

\frac{3}{2} e^{2 x} x^{2} - \frac{7}{2} e^{2 x} x + \frac{7}{4} e^{2 x} + C

Solution steps

\int (3 x^{2} - 4 x) e^{2 x} d x

Expand

(3 x^{2} - 4 x) e^{2 x} : 3 e^{2 x} x^{2} - 4 e^{2 x} x

= \int 3 e^{2 x} x^{2} - 4 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 3 e^{2 x} x^{2} d x - \int 4 e^{2 x} x d x

\int 3 e^{2 x} x^{2} d x = 3 (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x}))

\int 4 e^{2 x} x d x = 2 e^{2 x} x - e^{2 x}

= 3 (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})) - (2 e^{2 x} x - e^{2 x})

Simplify

3 (\frac{1}{2} e^{2 x} x^{2} - \frac{1}{4} (e^{2 x} \cdot 2 x - e^{2 x})) - (2 e^{2 x} x - e^{2 x}) : \frac{3}{2} e^{2 x} x^{2} - \frac{7}{2} e^{2 x} x + \frac{7}{4} e^{2 x}

= \frac{3}{2} e^{2 x} x^{2} - \frac{7}{2} e^{2 x} x + \frac{7}{4} e^{2 x}

Add a constant to the solution

= \frac{3}{2} e^{2 x} x^{2} - \frac{7}{2} e^{2 x} x + \frac{7}{4} e^{2 x} + C

n = 0 \sum \infty \frac{1}{n ^{2} + 3 n + 2}

x \to 0 lim (\frac{e ^{x} + e ^{- x}}{e ^{x} - e ^{- x}})

\frac{d}{d x} (1 - x)

(0, - 4), (1, 2)

x y^{^{'}} + 3 y = 3 e^{x}

What is the integral of (3x^2-4x)e^{2x} ?
The integral of (3x^2-4x)e^{2x} is 3/2 e^{2x}x^2-7/2 e^{2x}x+7/4 e^{2x}+C