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

The integral of x^2*e^{-4x} is -1/32 e^{-4x}(8x^2+4x+1)+C

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

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

- \frac{1}{32} e^{- 4 x} (8 x^{2} + 4 x + 1) + C

Solution steps

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

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{64} d u

= \int - \frac{e ^{u} u ^{2}}{64} d u

Take the constant out:

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

= - \frac{1}{64} \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

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

= - \frac{1}{64} (u^{2} e^{u} - \int 2 u e^{u} d u)

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

= - \frac{1}{64} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 4 x

= - \frac{1}{64} ((- 4 x)^{2} e^{- 4 x} - 2 (e^{- 4 x} (- 4 x) - e^{- 4 x}))

Simplify

- \frac{1}{64} ((- 4 x)^{2} e^{- 4 x} - 2 (e^{- 4 x} (- 4 x) - e^{- 4 x})) : - \frac{1}{32} e^{- 4 x} (8 x^{2} + 4 x + 1)

= - \frac{1}{32} e^{- 4 x} (8 x^{2} + 4 x + 1)

Add a constant to the solution

= - \frac{1}{32} e^{- 4 x} (8 x^{2} + 4 x + 1) + C

\int sec (4 x) tan (4 x) d x

y = \frac{1}{x - 3}

\frac{\partial}{\partial x} (x^{2} y^{3} - 1)

f (x) = (x^{4} + 2 x)^{3}, at x = 2

4 e^{x + 2} + e^{- 3}

What is the integral of x^2*e^{-4x} ?
The integral of x^2*e^{-4x} is -1/32 e^{-4x}(8x^2+4x+1)+C