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

The integral of x^3e^{-x/2} is 16(-1/8 e^{-x/2}x^3-3(1/4 e^{-x/2}x^2+e^{-x/2}x+2e^{-x/2}))+C

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

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

16 (- \frac{1}{8} e^{- \frac{x}{2}} x^{3} - 3 (\frac{1}{4} e^{- \frac{x}{2}} x^{2} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}})) + C

Solution steps

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

Apply u-substitution

: \int 16 e^{u} u^{3} d u

= \int 16 e^{u} u^{3} d u

Take the constant out:

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

= 16 \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

: u^{3} e^{u} - \int 3 u^{2} e^{u} d u

= 16 (u^{3} e^{u} - \int 3 u^{2} e^{u} d u)

\int 3 u^{2} e^{u} d u = 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

= 16 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - \frac{x}{2}

= 16 ((- \frac{x}{2})^{3} e^{- \frac{x}{2}} - 3 ((- \frac{x}{2})^{2} e^{- \frac{x}{2}} - 2 (e^{- \frac{x}{2}} (- \frac{x}{2}) - e^{- \frac{x}{2}})))

Simplify

16 ((- \frac{x}{2})^{3} e^{- \frac{x}{2}} - 3 ((- \frac{x}{2})^{2} e^{- \frac{x}{2}} - 2 (e^{- \frac{x}{2}} (- \frac{x}{2}) - e^{- \frac{x}{2}}))) : 16 (- \frac{1}{8} e^{- \frac{x}{2}} x^{3} - 3 (\frac{1}{4} e^{- \frac{x}{2}} x^{2} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}}))

= 16 (- \frac{1}{8} e^{- \frac{x}{2}} x^{3} - 3 (\frac{1}{4} e^{- \frac{x}{2}} x^{2} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}}))

Add a constant to the solution

= 16 (- \frac{1}{8} e^{- \frac{x}{2}} x^{3} - 3 (\frac{1}{4} e^{- \frac{x}{2}} x^{2} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}})) + C

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What is the integral of x^3e^{-x/2} ?
The integral of x^3e^{-x/2} is 16(-1/8 e^{-x/2}x^3-3(1/4 e^{-x/2}x^2+e^{-x/2}x+2e^{-x/2}))+C