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

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

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

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

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

Solution steps

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

Apply u-substitution

= \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

= - \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

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