How can one calculate this diferential equation $y\ dy = 4x(y^2+1)^2\ dx \text{ and } y(0) = 1$
I'm trying:
$\dfrac{y\ dy}{ (y^2+1)^2 }= 4x\ dx$
but I can't figure out what to do now.

Good start, you separated the variables. Now integrate. 
For the integral of the left-hand side, make the substitution $u=y^2+1$. You should arrive at 
$$\int \frac{1}{2} \cdot\frac{1}{u^2} \,du,$$
which is easy.
The integral of the right-hand side is $2x^2+C$. Use the initial condition to find $C$.
Added detail: Let $u=y^2+1$. Then $du =2y\,dy$ and therefore $y\,dy=\frac{1}{2} \,du$. Thus
$$\int \frac{y\,dy}{(1+y^2)^2}=\int \frac{1}{2}\cdot \frac{du}{u^2}=-\frac{1}{2}\cdot\frac{1}{u}.$$
(We deliberately and not quite correctly left out the constant of integration, since we will have one on the right, and one is enough.)
Thus an antiderivative of the expression on the left is $-\dfrac{1}{2(1+y^2)}$.
Integrating on the right, we end up with
$$ -\frac{1}{2(1+y^2)}=2x^2 +C.$$
Put $x=0$. Since $y(0)=1$, we get $C=-\frac{1}{4}$.
We have arrived at 
$$-\frac{1}{2(1+y^2)}=2x^2-\frac{1}{4}=\frac{8x^2-1}{4}.$$
Change sign and flip over. We get
$$2(1+y^2)=\frac{4}{1-8x^2.}$$
Now a small amount of manipulation gets us $y$ explicitly in terms of $x$. At the end we have to take a square root. Take the positive one, since $y(0)=1\gt 0$.
