What is the integral of (x^2-3x)e^{(-x)/3} ?

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

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integral of (x^2-3x)e^{(-x)/3}

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

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

Solution steps

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

Expand

(x^{2} - 3 x) e^{\frac{- x}{3}} : e^{- \frac{x}{3}} x^{2} - 3 e^{- \frac{x}{3}} x

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

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

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

\int e^{- \frac{x}{3}} x^{2} d x = - 3 e^{- \frac{x}{3}} x^{2} + 6 (- 3 e^{- \frac{x}{3}} x - 9 e^{- \frac{x}{3}})

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

= - 3 e^{- \frac{x}{3}} x^{2} + 6 (- 3 e^{- \frac{x}{3}} x - 9 e^{- \frac{x}{3}}) - 3 (- 3 e^{- \frac{x}{3}} x - 9 e^{- \frac{x}{3}})

Simplify

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

Add a constant to the solution

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

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