What is the integral of (x^3-1)e^{-x/3} ?

The integral of (x^3-1)e^{-x/3} is -3e^{-x/3}x^3-27e^{-x/3}x^2-162e^{-x/3}x-483e^{-x/3}+C

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integral of (x^3-1)e^{-x/3}

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

- 3 e^{- \frac{x}{3}} x^{3} - 27 e^{- \frac{x}{3}} x^{2} - 162 e^{- \frac{x}{3}} x - 483 e^{- \frac{x}{3}} + C

Solution steps

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

Expand

(x^{3} - 1) e^{- \frac{x}{3}} : e^{- \frac{x}{3}} x^{3} - e^{- \frac{x}{3}}

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

\int e^{- \frac{x}{3}} x^{3} d x = 81 (- \frac{1}{27} e^{- \frac{x}{3}} x^{3} - 3 (\frac{1}{9} e^{- \frac{x}{3}} x^{2} - 2 (- \frac{1}{3} e^{- \frac{x}{3}} x - e^{- \frac{x}{3}})))

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

= 81 (- \frac{1}{27} e^{- \frac{x}{3}} x^{3} - 3 (\frac{1}{9} e^{- \frac{x}{3}} x^{2} - 2 (- \frac{1}{3} e^{- \frac{x}{3}} x - e^{- \frac{x}{3}}))) - (- 3 e^{- \frac{x}{3}})

Simplify

81 (- \frac{1}{27} e^{- \frac{x}{3}} x^{3} - 3 (\frac{1}{9} e^{- \frac{x}{3}} x^{2} - 2 (- \frac{1}{3} e^{- \frac{x}{3}} x - e^{- \frac{x}{3}}))) - (- 3 e^{- \frac{x}{3}}) : - 3 e^{- \frac{x}{3}} x^{3} - 27 e^{- \frac{x}{3}} x^{2} - 162 e^{- \frac{x}{3}} x - 483 e^{- \frac{x}{3}}

= - 3 e^{- \frac{x}{3}} x^{3} - 27 e^{- \frac{x}{3}} x^{2} - 162 e^{- \frac{x}{3}} x - 483 e^{- \frac{x}{3}}

Add a constant to the solution

= - 3 e^{- \frac{x}{3}} x^{3} - 27 e^{- \frac{x}{3}} x^{2} - 162 e^{- \frac{x}{3}} x - 483 e^{- \frac{x}{3}} + C

\frac{d}{d x} (- \frac{120}{x ^{6}})

\frac{d}{d z} (x y)

x^{3} y^{^{'}} + 2 x^{2} y = - 4 cos (2 x) y^{- \frac{1}{2}}

x \to 1 lim (\frac{5}{x ^{3} - 1})

n = 0 \sum \infty 4 n^{2}

What is the integral of (x^3-1)e^{-x/3} ?
The integral of (x^3-1)e^{-x/3} is -3e^{-x/3}x^3-27e^{-x/3}x^2-162e^{-x/3}x-483e^{-x/3}+C