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

The integral of x^3e^{-x/3} is 81(-1/27 e^{-x/3}x^3-3(1/9 e^{-x/3}x^2-2(-1/3 e^{-x/3}x-e^{-x/3})))+C

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

\int x^{3} e^{- \frac{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}}))) + C

Solution steps

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

Apply u-substitution

: \int 81 e^{u} u^{3} d u

= \int 81 e^{u} u^{3} d u

Take the constant out:

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

= 81 \cdot \int e^{u} u^{3} d u

Apply Integration By Parts

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

= 81 (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}))

= 81 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - \frac{x}{3}

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

Simplify

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

= 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}})))

Add a constant to the solution

= 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}}))) + C

x^{4}, x

\frac{\partial}{\partial y} (ln (y))

\frac{\frac{E}{L}}{( s ^{2} + \frac{1}{C L} )}

y = x^{π}

f (x) = \frac{x}{( 7 x - 6 ) ^{6}}, at x = 1

What is the integral of x^3e^{-x/3} ?
The integral of x^3e^{-x/3} is 81(-1/27 e^{-x/3}x^3-3(1/9 e^{-x/3}x^2-2(-1/3 e^{-x/3}x-e^{-x/3})))+C