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January 30, 2015

January 30, 2015

Posted by **Jay** on Monday, December 11, 2006 at 12:40pm.

(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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