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

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