Question

Evaluate using integration by parts ​

Answer 1

Rather than carrying out IBP several times, let's establish a more general result. Let

`I(n)=\displaystyle\int x^ne^x\,\mathrm dx`

One round of IBP, setting

`u=x^n\implies\mathrm du=nx^(n-1)\,\mathrm dx`

`\mathrm dv=e^x\,\mathrm dx\implies v=e^x`

gives

`\displaystyle I(n)=x^ne^x-n\int x^(n-1)e^x\,\mathrm dx`

`I(n)=x^ne^x-nI(n-1)`

This is called a power-reduction formula. We could try solving for
`I(n)` explicitly, but no need.
`n=5` is small enough to just expand
`I(5)` as much as we need to.

`I(5)=x^5e^x-5I(4)`

`I(5)=x^5e^x-5(x^4e^x-4I(3))=(x-5)x^4e^x+20I(3)`

`I(5)=(x-5)x^4e^x+20(x^3e^x-3I(2))=(x^2-5x+20)x^3e^x-60I(2)`

`I(5)=(x^2-5x+20)x^3e^x-60(x^2e^x-2I(1))=(x^3-5x^2+20x-60)x^2e^x+120I(1)`

`I(5)=(x^3-5x^2+20x-60)x^2e^x+120(xe^x-I(0))`

Finally,

`I(0)=\displaystyle\int e^x\,\mathrm dx=e^x+C`

so we end up with

`I(5)=(x^4-5x^3+20x^2-60x+120)xe^x-120e^x+C`

`I(5)=(x^5-5x^4+20x^3-60x^2+120x-120)e^x+C`

and the antiderivative is

`\displaystyle\int2x^5e^x\,\mathrm dx=(2x^5-10x^4+40x^3-120x^2+240x-240)e^x+C`