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

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

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

integral

x^{3} \cdot (e^{- x})^{2}

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

Solution steps

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

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

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

Apply u-substitution

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

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

Take the constant out:

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

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

Substitute back

u = - 2 x

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

Simplify

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

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

Add a constant to the solution

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

\frac{x ^{2}}{( x ^{3} + 1 ) ^{\frac{1}{2}}}

\frac{- 2}{e ^{x}}

\frac{1}{( x ( x ^{2} + x + 1 ) )}

5 (1 - x)^{4}

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

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