What is the integral of 3x^2e^{2x} ?

The integral of 3x^2e^{2x} is 3(1/2 e^{2x}x^2-1/4 (e^{2x}*2x-e^{2x}))+C

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integral of 3x^2e^{2x}

\int 3 x^{2} e^{2 x} d x

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

Solution steps

\int 3 x^{2} e^{2 x} d x

Take the constant out:

\int a \cdot f (x) d x = a \cdot \int f (x) d x

= 3 \cdot \int x^{2} e^{2 x} d x

Apply Integration By Parts

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

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

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

Add a constant to the solution

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

n = 1 \sum \infty \frac{1}{n} \cdot x^{n}

\frac{d}{d x} (y x)

y^{^{''}} - 3 y^{^{'}} + 2 y = 0, y (0) = 1, y^{^{'}} (0) = 0

y = \frac{- 6 x}{x ^{2} + 1}, (1, - 3)

y^{^{'}} = y (1 - 0.001 y)

What is the integral of 3x^2e^{2x} ?
The integral of 3x^2e^{2x} is 3(1/2 e^{2x}x^2-1/4 (e^{2x}*2x-e^{2x}))+C