What is the integral of x^4*(e^{-x})^2 ?

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

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integral of x^4*(e^{-x})^2

integral

x^{4} \cdot (e^{- x})^{2}

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

Solution steps

\int x^{4} (e^{- x})^{2} d x

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

= \int e^{- 2 x} x^{4} d x

Apply u-substitution

: \int - \frac{e ^{u} u ^{4}}{32} d u

= \int - \frac{e ^{u} u ^{4}}{32} d u

Take the constant out:

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

= - \frac{1}{32} \cdot \int e^{u} u^{4} d u

Apply Integration By Parts

: u^{4} e^{u} - \int 4 u^{3} e^{u} d u

= - \frac{1}{32} (u^{4} e^{u} - \int 4 u^{3} e^{u} d u)

\int 4 u^{3} e^{u} d u = 4 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u})))

= - \frac{1}{32} (u^{4} e^{u} - 4 (u^{3} e^{u} - 3 (u^{2} e^{u} - 2 (e^{u} u - e^{u}))))

Substitute back

u = - 2 x

= - \frac{1}{32} ((- 2 x)^{4} e^{- 2 x} - 4 ((- 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}{32} ((- 2 x)^{4} e^{- 2 x} - 4 ((- 2 x)^{3} e^{- 2 x} - 3 ((- 2 x)^{2} e^{- 2 x} - 2 (e^{- 2 x} (- 2 x) - e^{- 2 x})))) : - \frac{1}{4} e^{- 2 x} (2 x^{4} + 4 x^{3} + 3 (2 x^{2} + 2 x + 1))

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

Add a constant to the solution

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

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

x^{7} - 6 x + 8

\frac{1}{( e ^{m} )}

\frac{x ^{2} + x ^{4}}{4 x}

- x + 6

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