attempting to solve the diff equation $y'(x^2+1)-2xy=4\sqrt{y(x^2+1)}$ $$y'(x^2+1)-2xy=4\sqrt{y(x^2+1)}$$
$$y'\sqrt{(x^2+1)}-\frac{2xy}{\sqrt{(x^2+1)}}=4\sqrt{y}$$
$$\frac{y'\sqrt{(x^2+1)}-\frac{2xy}{\sqrt{(x^2+1)}}}{x^2+1}= \frac{4\sqrt{y}}{x^2+1}$$
I tried to solve this but I can't get rid of the 2 which would have enabled me to wrtie $$(\frac{y}{x^2+1})'$$
how do I solve this? 

It's Bernouilli's equation
$$y'(x^2+1)-2xy=4\sqrt{y(x^2+1)}$$
do this rather
$$y'y^{-1/2}(x^2+1)-2xy^{1/2}=4\sqrt{(x^2+1)}$$
$$2(y^{1/2})'(x^2+1)-2x(y^{1/2})=4\sqrt{(x^2+1)}$$
Substitute $z=y^{1/2}$
$$z'(x^2+1)-xz=2\sqrt{(x^2+1)}$$
Now you can solve it easily...

Edit
$$(\frac{y}{x^2+1})'= \frac{4\sqrt{y}}{x^2+1}$$
You could do this starting from where you were stuck
$$(\frac{y}{x^2+1})'= 4\sqrt{\frac{{y}}{x^2+1}}\frac{1}{\sqrt{x^2+1}}$$
$$\frac {d(\frac{y}{x^2+1})}{\sqrt{\frac{{y}}{x^2+1}}}= \frac{4dx}{\sqrt{x^2+1}}$$
And simply integrate...
$$\sqrt{\frac{{y}}{x^2+1}}= 2 \int \frac{dx}{\sqrt{x^2+1}}$$
But Bernouilli's method is easier
