How do I find the integral of (64x+8)e^(8x^2+2x+1)dx

Use substitution.

Let u = 8x² + 2x + 1. Then
du = (16x + 2)dx
4 du = (64x + 8) dx

and the integral becomes

∫ 4e^u du

which of course equals 4e^u + C or, in terms of the original variable x,
4e^(8x² + 2x + 1) + C.