What is the integral of 2x^2e^{3x+1} ?

The integral of 2x^2e^{3x+1} is 2e(1/3 e^{3x}x^2-2/27 (3e^{3x}x-e^{3x}))+C

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integral of 2x^2e^{3x+1}

\int 2 x^{2} e^{3 x + 1} d x

2 e (\frac{1}{3} e^{3 x} x^{2} - \frac{2}{27} (3 e^{3 x} x - e^{3 x})) + C

Solution steps

\int 2 x^{2} e^{3 x + 1} d x

Take the constant out:

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

= 2 \cdot \int x^{2} e^{3 x + 1} d x

x^{2} e^{3 x + 1} = e^{1} x^{2} e^{3 x}

= 2 \cdot \int e^{1} x^{2} e^{3 x} d x

Take the constant out:

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

= 2 e^{1} \cdot \int x^{2} e^{3 x} d x

Simplify

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

Apply Integration By Parts

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

= 2 e (\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})

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

Add a constant to the solution

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

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What is the integral of 2x^2e^{3x+1} ?
The integral of 2x^2e^{3x+1} is 2e(1/3 e^{3x}x^2-2/27 (3e^{3x}x-e^{3x}))+C