# Math

Find the integrals. (show steps)

(integral sign) xe^(4x^2)

I think this how is how its done:

(integral sign) xe^(4x^2)
it's a u du problem
let u=4x^2
so, du=8x dx

now you have an x already so all u need is 8 inside and and 1/8 outside the integral

[1/8] (integral sign) [8]xe^(4x^2) dx
1/8(integral sign) e^u du
1/8 e^(4x^2) + C
DONE

Ah. I should have known. Thanks

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