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

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

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

integral

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

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

Solution steps

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

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

= \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

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

Simplify

= 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

= 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})

= 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

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

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

e^{8} x cos (x)

(5 - 3 x) (2 + x^{2})

(d^{4} - 2 d^{2} + 1) y = cos (x)

(d^{2} - 4 d + 4) y = e^{2 x} + cos (2 x) + x^{3}

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