Convergence of $\int_{-1}^{1} \frac{dx}{x\sqrt{1-x^2}}$ I was supposed to calculate integral
$$
I=\int_{-1}^{1} \frac{dx}{x\sqrt{1-x^2}}
$$
My teacher gave my following solution:
$$
I=\sum_{i=-2}^1 \int_{\frac{i}{2}}^{\frac{i+1}{2}} \frac{dx}{x\sqrt{1-x^2}}
$$
and for example:
$$
\forall _{x\in(0;\frac{1}{2}]} \frac{1}{x\sqrt{1-x^2}}\geq\frac{1}{x}>0
$$
therefore:
$$
I_3=\int_{0}^{\frac{1}{2}} \frac{dx}{x\sqrt{1-x^2}}
$$
is divergent, so $I$ is divergent.
Is this a valid solution? Proving that one of the fragments of the integral is divergent implies that the whole integral is divergent? I think it's an odd function, and since the integration intervals are symmetrical, then the integral should be convergent and equal $0$.
 A: You may not apply symmetric arguments as the domain of integration contains three points where the function has discontinuities, specifically at x=-1,0,1. Ideally, the first step in solving these improper integrals is to break it in three different pieces with each domain containing one of the discontinuities, and then study each integral separately. If one can find that one of the integrals diverge, then the entire integral surely diverge.
As was pointed out in the comments, you teacher's solution employs unconventional notation, but, in fact, it is not about the notation, but the procedure. I believe theres no need to break the integrals in 5 pieces, and the show through a lousy argument that one of the pieces diverges thus the entire integral diverges. It is just a hand-weavy way to do things, that I believe it may be great to teach when you already know the material well and you can allow certain sloppiness, but if you are a beginner, youre better off learning it the correct way and not the slick way 
