# calculus

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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)/(1-x^2)

• calculus -

dy = (3x)/(1-x^2) dx
y = Integral of [3x/(1-x^2)] dx + constant

For the integration, let
1-x^2 = u
-2x dx = du
Which make the integral
Integral of (-3/2)du/u = -(3/2)ln(1-x^2)

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