# Integrate this how? $\int {\frac {y(x)' (1-y(x)^2)}{y(x)^3}} dx$

I'm trying to solve a differential equation but I get to the point in which I have to solve this:
$$\int {\frac {y(x)' (1-y(x)^2)}{y(x)^3}} dx$$
I don't understand how to do it. I think it's related to some property I don't know, but the $y(x)'$ is throwing me off. I tried looking at Wolfram's step by step but I don't understand it.
Thanks for any help.

Context: Original function to solve is $(v^3) du + (u^3-uv^2) dv = 0$.
I used a variable change: $y = \dfrac uv$. I'm not sure if it's correct or anything though.

• What is the differential equation you're trying to solve? How does this integral come up? If you're using separation of variable there shouldn't be a $y'$ in any integral. – DanZimm Jun 6 '14 at 6:26
• Hint: use the chain rule. – David H Jun 6 '14 at 6:27
• Also welcome to MSE! – DanZimm Jun 6 '14 at 6:27
• @DavidH Oh, I always forget the chain rule for integrals. Thanks! – Paula Jun 6 '14 at 6:32
• And @DanZimm I used first a change of variables, edited the question for context if you're curious :). And thanks for the welcome! – Paula Jun 6 '14 at 6:33

Using the obvious change of variables $y=y(x) \implies dy=y'(x)\,dx$, your integral becomes:
$$\int {\frac {y(x)' (1-y(x)^2)}{y(x)^3}} dx=\int {\frac {1-y^2}{y^3}} dy\\ =\int y^{-3}dy-\int y^{-1}dy\\ =-\frac12 y^{-2}-\log{y}+constant$$
The differential equation simplifies a lot if you use $$u(v)=\frac{v}{y(v)}$$ (which is basically the reciprocal of what you chose).
After replacement and simplifications (there are several), you should arrive to $$1-v y(v) \frac{dy(v)}{dv}=0$$ for which the solution is simply $$y(v)= \sqrt{c_1+2\log (v)}$$ from which $$u(v)=\frac{v}{\sqrt{c_1+2 \log (v)}}$$