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

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

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

integral

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

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

Solution steps

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

Take the constant out:

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

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

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

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

Apply u-substitution

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

= - 2 \cdot \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

= - 2 \cdot \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

= - 2 \cdot \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}))

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

Substitute back

u = - 2 x

= - 2 \cdot \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

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

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

Add a constant to the solution

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

\frac{4 + 5 sin ( x )}{cos ( x )}

\frac{1}{( 3 x )}

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

f (x)^{2}

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

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