How do you integrate $(x^2)(e^x)dx$ ?

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Answer 1 · 2016-09-17T09:41:14

$intx^2e^xdx=e^x(x^2-2x+2)+c$

Let $u=x^2$ and $v=e^x$ , then $du=2xdx$ and $dv=e^xdx$

Now integration by parts states that

$intu(x)v'(x)dx=u(x)v(x)-intv(x)u'(x)dx$

Hence $intx^2e^xdx=x^2e^x-inte^x xx 2xdx$

= $x^2e^x-2intxe^xdx+c$ ...............(1)

Now we set $u=x$ , then $du=dx$

and $intxe^xdx=xe^x-inte^x xx1xxdx$ or

$intxe^xdx=xe^x-inte^xdx=xe^x-e^x$

Putting this in (1), we get

$intx^2e^xdx=x^2e^x-2(xe^x-e^x)+c$

= $e^x(x^2-2x+2)+c$

How do you integrate (x^2)(e^x)dx(x^2)(e^x)dx?