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

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

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

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

- e^{- 3 x} x^{2} - \frac{2}{3} e^{- 3 x} x - \frac{2}{9} e^{- 3 x} + C

Solution steps

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

Take the constant out:

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

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

Apply u-substitution

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

= 3 \cdot \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

= 3 (- \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

= 3 (- \frac{1}{27} (u^{2} e^{u} - \int 2 u e^{u} d u))

Take the constant out:

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

= 3 (- \frac{1}{27} (u^{2} e^{u} - 2 \cdot \int u e^{u} d u))

Apply Integration By Parts

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

= 3 (- \frac{1}{27} (u^{2} e^{u} - 2 (e^{u} u - \int e^{u} d u)))

Use the common integral:

\int e^{u} d u = e^{u}

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

Substitute back

u = - 3 x

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

Simplify

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

= - e^{- 3 x} x^{2} - \frac{2}{3} e^{- 3 x} x - \frac{2}{9} e^{- 3 x}

Add a constant to the solution

= - e^{- 3 x} x^{2} - \frac{2}{3} e^{- 3 x} x - \frac{2}{9} e^{- 3 x} + C

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