Question

How do you integrate `int x^2e^-x` by integration by parts method?

Answer 1

`= - e^-x ( x^2 + 2x + 2) + C`

Explanation:

`int \ x^2e^-x \ dx`

`= int \ x^2d/dx( - e^-x) \ dx`

which by IBP means
`= - x^2 e^-x - int \ - e^-x d/dx(x^2 ) \ dx`

`= - x^2 e^-x + int \ e^-x * 2x\ dx`

setting up one more round of IBP

`= - x^2 e^-x + int \ d/dx( - e^-x) * 2x\ dx`

`= - x^2 e^-x + ( - e^-x)* 2x - int \ - e^-x * d/dx( 2x)\ dx`

`= - x^2 e^-x - 2x e^-x + 2 int \ e^-x\ dx`

`= - x^2 e^-x - 2x e^-x - 2 e^-x + C`

`= - e^-x ( x^2 + 2x + 2) + C`

Answer 2

`-e^{-x}(x^2+2x+2+C_1)`

Explanation:

`d/(dx)(x^n e^{-x})=nx^{n-1}e^{-x}-x^n e^{-x}`

Calling `I_n = int x^n e^{-x}dx` we have

`I_n-nI_{n-1}=-x^n e^{-x}`

making now `J_n = e^xI_n` we obtain a simpler recurrence relation

`J_n-nJ_{n-1} = -x^n`

This recurrence formulation allows us to obtain the general solution for the integral

`int x^n e^{-x}dx`

Now doing `n=2` we have

`{(J_2-2J_1=-x^2), (J_1-J_0 = -x) :}`

Multiplying the second equation by `2` and adding with the first we obtain

`J_2-2J_0=-x^2-2x`

but `J_0 = e^xI_0 = e^x int e^{-x}dx = -1+C`

so

`J_2 = e^x I_2 = -x^2-2x-2+2C`

Finally

`I_2 = -e^{-x}(x^2+2x+2+C_1)`