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

The integral of-(x^3)/(3e^{x/2)} is -2/3 (-e^{-x/2}x^3-24((e^{-x/2}x^2)/4+e^{-x/2}x+2e^{-x/2}))+C

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

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

- \frac{2}{3} (- e^{- \frac{x}{2}} x^{3} - 24 (\frac{e ^{- \frac{x}{2}} x ^{2}}{4} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}})) + C

Solution steps

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

Take the constant out:

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

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

Apply exponent rule:

\frac{1}{a ^{b}} = a^{- b}

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

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

Apply u-substitution

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

= - \frac{1}{3} \cdot \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

= - \frac{1}{3} \cdot 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

= - \frac{1}{3} \cdot 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}))

= - \frac{1}{3} \cdot 16 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

Substitute back

u = - \frac{x}{2}

= - \frac{1}{3} \cdot 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

- \frac{1}{3} \cdot 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}}))) : - \frac{2}{3} (- e^{- \frac{x}{2}} x^{3} - 24 (\frac{e ^{- \frac{x}{2}} x ^{2}}{4} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}}))

= - \frac{2}{3} (- e^{- \frac{x}{2}} x^{3} - 24 (\frac{e ^{- \frac{x}{2}} x ^{2}}{4} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}}))

Add a constant to the solution

= - \frac{2}{3} (- e^{- \frac{x}{2}} x^{3} - 24 (\frac{e ^{- \frac{x}{2}} x ^{2}}{4} + e^{- \frac{x}{2}} x + 2 e^{- \frac{x}{2}})) + C

\frac{d}{d q} (500 q - \frac{q ^{2}}{ln ( q + 300 )})

\frac{\partial}{\partial x} (tan (9 + 2 x^{2} y^{4} z^{2}))

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

\frac{1}{8} x, \frac{1}{x ^{2}}, x

x \to \infty lim (ln (5 x^{2} + 12 x - 6))

What is the integral of-(x^3)/(3e^{x/2)} ?
The integral of-(x^3)/(3e^{x/2)} is -2/3 (-e^{-x/2}x^3-24((e^{-x/2}x^2)/4+e^{-x/2}x+2e^{-x/2}))+C