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

The integral of x^2e^{-3x+1} is -1/3 e^{-3x+1}x^2+2/3 (-1/3 e^{-3x+1}x-1/9 e^{-3x+1})+C

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

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

- \frac{1}{3} e^{- 3 x + 1} x^{2} + \frac{2}{3} (- \frac{1}{3} e^{- 3 x + 1} x - \frac{1}{9} e^{- 3 x + 1}) + C

Solution steps

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

Apply Integration By Parts

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

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

\int - \frac{2}{3} e^{- 3 x + 1} x d x = - \frac{2}{3} (- \frac{1}{3} e^{- 3 x + 1} x - \frac{1}{9} e^{- 3 x + 1})

= - \frac{1}{3} e^{- 3 x + 1} x^{2} - (- \frac{2}{3} (- \frac{1}{3} e^{- 3 x + 1} x - \frac{1}{9} e^{- 3 x + 1}))

Simplify

= - \frac{1}{3} e^{- 3 x + 1} x^{2} + \frac{2}{3} (- \frac{1}{3} e^{- 3 x + 1} x - \frac{1}{9} e^{- 3 x + 1})

Add a constant to the solution

= - \frac{1}{3} e^{- 3 x + 1} x^{2} + \frac{2}{3} (- \frac{1}{3} e^{- 3 x + 1} x - \frac{1}{9} e^{- 3 x + 1}) + C

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What is the integral of x^2e^{-3x+1} ?
The integral of x^2e^{-3x+1} is -1/3 e^{-3x+1}x^2+2/3 (-1/3 e^{-3x+1}x-1/9 e^{-3x+1})+C