Simple first order differential equation I've got another first order differential that has me stumped.  Please see my work and let me know where I'm going wrong:
$$y' = \frac{6x^2}{y(1+x^3)}$$
$$y'y = \frac{6x^2}{1+x^3}$$
$$ y.\frac{dy}{ dx} = \frac{6x^2}{1+x^3} $$
$$\int y dy = \int 6x^2\frac{1}{1+x^3} dx$$
$$\frac{1}{2}y^2 = 2x^3 \ln|1 + x^3| + c$$
$$y^2 = 4x^3 \ln|1 + x^3| + c$$
$$y = \pm \sqrt{4x^3 \ln |1 + x^3| + c}$$
Unfortunately, my answer is wrong and I'm not quite sure why.  Please help.
 A: Your error is that $\displaystyle\int\dfrac{6x^2}{1+x^3}\,dx = 2\ln|1+x^3|+C$ and not $2x^3\ln|1+x^3|+C$. 
A: Unlike the integral of a sum or difference of functions , where $$\int (f(x) \pm g(x))\,dx = \int f(x) \,dx \pm \int g(x)\,dx,$$ the integral of the product of functions is not the product of the integrals of each function.  I.e., $$\int f(x)g(x)\,dx \neq \left(\int f(x)\,dx\right)\cdot \left(\int g(x) \,dx\right)$$
And in any case, $\displaystyle \int \dfrac 1{1+x^3}\,dx \neq \ln|1 + x^3|+C$, because $$u = 1+x^3 \iff x = (u-1)^{1/3} \implies  dx = \frac 13(u-1)^{-2/3}\,du$$ to give us $$\int \dfrac 1{1+x^3}\,dx = \frac 13 \int \frac{du}{u(u-1)^{2/3}}\neq \ln|u| + C = \ln|1 + x^3| + C$$

Okay, so lets take a different direction. Note that if $u = (1+x^3), \;du = 3x^2\,dx.\;$ We see that $du$ is very close to our numerator, which is $6x^2 = 2\cdot 3x^2$.
So $$\int \dfrac{6x^2}{1+x^3}\,dx =2\int \dfrac{3x^2\,dx}{1+x^3} =2\int \dfrac{du}{u}= 2 \ln|u| + C$$ $$ =2\ln|1+x^3| + C, \text{ or } \ln((1+x^3)^2) + C$$
