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

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

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

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

- 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^{- 2 x} 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^{- 2 x} 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

x \to \frac{π}{4} lim (\frac{cos ( 2 x )}{2 cos ( x ) - 1})

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\frac{d}{d x} (6 ln (x^{2}))

\int x \cdot tan^{2} (x) d x

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