calculus
posted by Annie .
I am having alot of trouble remembering how to do this and this is a practice question for my final if you could help that would be great. Solve the differential equation (dy/dx)=(3x)/(1x^2)

dy = (3x)/(1x^2) dx
y = Integral of [3x/(1x^2)] dx + constant
For the integration, let
1x^2 = u
2x dx = du
Which make the integral
Integral of (3/2)du/u = (3/2)ln(1x^2)
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