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

The integral of x^2e^{-ax} is -e^{-ax} 1/a x^2+2/(a^3)(-ae^{-ax}x-e^{-ax})+C

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integral of x^2e^{-ax}

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

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

Solution steps

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

Apply Integration By Parts

: - e^{- a x} \frac{1}{a} x^{2} - \int - e^{- a x} \frac{2}{a} x d x

= - e^{- a x} \frac{1}{a} x^{2} - \int - e^{- a x} \frac{2}{a} x d x

\int - e^{- a x} \frac{2}{a} x d x = - \frac{2}{a ^{3}} (- a e^{- a x} x - e^{- a x})

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

Simplify

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

Add a constant to the solution

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

\int cos (\frac{x}{9}) d x

(3, - 29), (- 1, 7)

y^{^{''}} + 6 y^{^{'}} + 9 y = e^{- 3 t}

\int_{1}^{5} \frac{ln ( x )}{x} d x

\int_{- \infty}^{0} \frac{1}{5 - 9 x} d x

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