What is the integral of (x^3)/(e^x) ?

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

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integral of (x^3)/(e^x)

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

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

Solution steps

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

Apply exponent rule:

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

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

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

Apply u-substitution

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

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

Apply Integration By Parts

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

= 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}))

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

Substitute back

u = - x

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

Simplify

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

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

Add a constant to the solution

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

\frac{d}{d x} (\frac{4 x ^{5} + 6}{x ^{3}})

- \frac{5}{x}

x, x^{2}, [0, 1]

x \to \infty lim (x \cdot 3)

x + y = 5

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