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

The integral of e^{-3x}x^2 is -1/27 (9e^{-3x}x^2-2(-3e^{-3x}x-e^{-3x}))+C

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

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

- \frac{1}{27} (9 e^{- 3 x} x^{2} - 2 (- 3 e^{- 3 x} x - e^{- 3 x})) + C

Solution steps

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

Apply u-substitution

: \int - \frac{e ^{u} u ^{2}}{27} d u

= \int - \frac{e ^{u} u ^{2}}{27} d u

Take the constant out:

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

= - \frac{1}{27} \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}{27} (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}{27} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 3 x

= - \frac{1}{27} ((- 3 x)^{2} e^{- 3 x} - 2 (e^{- 3 x} (- 3 x) - e^{- 3 x}))

Simplify

- \frac{1}{27} ((- 3 x)^{2} e^{- 3 x} - 2 (e^{- 3 x} (- 3 x) - e^{- 3 x})) : - \frac{1}{27} (9 e^{- 3 x} x^{2} - 2 (- 3 e^{- 3 x} x - e^{- 3 x}))

= - \frac{1}{27} (9 e^{- 3 x} x^{2} - 2 (- 3 e^{- 3 x} x - e^{- 3 x}))

Add a constant to the solution

= - \frac{1}{27} (9 e^{- 3 x} x^{2} - 2 (- 3 e^{- 3 x} x - e^{- 3 x})) + C

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What is the integral of e^{-3x}x^2 ?
The integral of e^{-3x}x^2 is -1/27 (9e^{-3x}x^2-2(-3e^{-3x}x-e^{-3x}))+C