Question #168566

y=x^2e^x find the indefinite integral funtion



1
Expert's answer
2021-03-04T10:40:17-0500

Let's integrate x2exdx\int x^2e^xdx by parts. Let's use the rule:


fg=fgfg,\int fg'=fg-\int f'g,

here, f=x2,g=exf=x^2, g=e^x.

Then, we get:


x2exdx=x2ex2xexdx.\int x^2e^xdx=x^2e^x-2\int xe^xdx.

Let's integrate xexdx\int xe^xdx by parts and use the same rule. This time f=x,g=exf=x, g=e^x. Finally, we get:


x2exdx=x2ex2(xexexdx),\int x^2e^xdx=x^2e^x-2(xe^x-\int e^xdx),x2exdx=x2ex2xex+2ex=(x22x+2)ex+C.\int x^2e^xdx=x^2e^x-2xe^x+2e^x=(x^2-2x+2)e^x+C.

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