What is the integral of x^2e^{-6x} ?

The integral of x^2e^{-6x} is -1/108 e^{-6x}(18x^2+6x+1)+C

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integral of x^2e^{-6x}

\int x^{2} e^{- 6 x} d x

- \frac{1}{108} e^{- 6 x} (18 x^{2} + 6 x + 1) + C

Solution steps

\int x^{2} e^{- 6 x} d x

Apply u-substitution

= \int - \frac{e ^{u} u ^{2}}{216} d u

Take the constant out:

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

= - \frac{1}{216} \cdot \int e^{u} u^{2} d u

Apply Integration By Parts

= - \frac{1}{216} (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}{216} (u^{2} e^{u} - 2 (e^{u} u - e^{u}))

Substitute back

u = - 6 x

= - \frac{1}{216} ((- 6 x)^{2} e^{- 6 x} - 2 (e^{- 6 x} (- 6 x) - e^{- 6 x}))

Simplify

- \frac{1}{216} ((- 6 x)^{2} e^{- 6 x} - 2 (e^{- 6 x} (- 6 x) - e^{- 6 x})) : - \frac{1}{108} e^{- 6 x} (18 x^{2} + 6 x + 1)

= - \frac{1}{108} e^{- 6 x} (18 x^{2} + 6 x + 1)

Add a constant to the solution

= - \frac{1}{108} e^{- 6 x} (18 x^{2} + 6 x + 1) + C

t a y l or e x^{2}

\int_{- 2}^{2} (x^{2} - 4) d x

x \to 0 lim (\frac{( 1 + x ) ^{4} - 1}{x})

in v erse l a pl a ce \frac{- 3 s + 9}{( s + 3 ) ^{2}}

\frac{d}{d t} (4 e^{t} sin (t))

What is the integral of x^2e^{-6x} ?
The integral of x^2e^{-6x} is -1/108 e^{-6x}(18x^2+6x+1)+C