IVP to $y'=\frac{2xy^2+2x}{x^2+1}$ Find a solution of  $y'=\frac{2xy^2+2x}{x^2+1}$ with given IVP $y(0)=\sqrt{3}$.
My solution: 
$\int \frac{1}{y^2+1}dy=\int \frac{2x}{x^2+1}dx$  $\Rightarrow \tan^{-1}(y)=\log(x^2+1)+c, c\in \mathbb{R}$  $\Rightarrow y= \tan (\log(x^2+1)+c),c\in \mathbb{R} $
$y(o)=\sqrt{3}=\tan(c)$  $ \Rightarrow c=\frac{4 \pi}{3}$
$\Rightarrow y= \tan (\log(x^2+1)+\frac{4 \pi}{3})$
Determining the domain of defeniton: 
$\frac{\pi}{2}= (\log(x^2+1)+\frac{4 \pi}{3})$ $ \Rightarrow x= \pm \sqrt{ e^{ \frac{-5\pi}{6}}-1}$ 
$-\frac{\pi}{2}=(\log(x^2+1)+\frac{4 \pi}{3})$ $\Rightarrow x=\pm \sqrt{ e^{ \frac{-11\pi}{6}}-1}$
So domain of definition would be  $D=]-\sqrt{ e^{ \frac{-11\pi}{6}}-1}, \sqrt{ e^{ \frac{-5\pi}{6}}-1}[$
Is this correct?

The solution and the steps leading up to them are all correct. You can always check by plugging the expression you obtained back into the differential equation. Using the chain rule, the derivatives of elementary functions, and trigonometric identities, you should find that both sides are equal to
$
\dfrac{2x \sec^2\left(\log(x^2+1)+\frac{4\pi}{3}\right)}{x^2 + 1}.
$
The domain of definition, on the other hand, isn't correct. Notice that the ends you gave for the maximal interval of existence are square roots of negative numbers, since $e^x < 1$ for all $x < 0$. The core issue is that $\log(x^2 + 1) + \frac{4\pi}{3} \geq \frac{4\pi}{3} > \pm \frac{\pi}{2}$, and so the two equations you were trying to solve in fact have no real solutions. In other words, the singularities of $\tan$ at $\pm\frac{\pi}{2}$ do not lead to singularities of your expression for $y$.
But $\pm\frac{\pi}{2}$ aren't the only forbidden inputs for $\tan$. Your expression for $y(x)$ blows up to $\pm\infty$ at $x$ wherever $\tan$ has a singularity at $\log(x^2+1) + \frac{4\pi}{3}$. What you need to do is to find the closest such $x$'s to the left and to the right of $0$.
