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

The integral of 3x^2ye^{3x} is ye^{3x}x^2-2/3 ye^{3x}x+2/9 ye^{3x}+C

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

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

y e^{3 x} x^{2} - \frac{2}{3} y e^{3 x} x + \frac{2}{9} y e^{3 x} + C

Solution steps

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

Take the constant out:

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

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

Apply Integration By Parts

: \frac{1}{3} e^{3 x} x^{2} - \int \frac{2}{3} e^{3 x} x d x

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

\int \frac{2}{3} e^{3 x} x d x = \frac{2}{27} (3 e^{3 x} x - e^{3 x})

= 3 y (\frac{1}{3} e^{3 x} x^{2} - \frac{2}{27} (3 e^{3 x} x - e^{3 x}))

Simplify

3 y (\frac{1}{3} e^{3 x} x^{2} - \frac{2}{27} (3 e^{3 x} x - e^{3 x})) : y e^{3 x} x^{2} - \frac{2}{3} y e^{3 x} x + \frac{2}{9} y e^{3 x}

= y e^{3 x} x^{2} - \frac{2}{3} y e^{3 x} x + \frac{2}{9} y e^{3 x}

Add a constant to the solution

= y e^{3 x} x^{2} - \frac{2}{3} y e^{3 x} x + \frac{2}{9} y e^{3 x} + C

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What is the integral of 3x^2ye^{3x} ?
The integral of 3x^2ye^{3x} is ye^{3x}x^2-2/3 ye^{3x}x+2/9 ye^{3x}+C