Question

what is its integral​

Answer 1

Answer:

-2x to the power 1 .

I think so.

Answer 2

Step-by-step explanation:

∫e^x(x−1)x2 dx

=∫e^x(1/x−1/x^2) dx

=∫ex1x dx−∫e^x 1x^2 dx

=1/x∫e^x dx−∫(dd/x * 1/x⋅∫e^x dx) dx−∫e^x * 1/x2 dx

=1x(e^x)−∫(−1/x^2 *e^x) dx−∫e^x * 1/x^2 dx

=e^x/x+∫e^x*1/x^2 dx−∫e^x *1/x^2 dx+C

=e^x/x+C

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