Is $\int_{-2}^3\frac{1}{x^3}dx=\frac5{72}$ or not defined? If we divide it into two parts such that $$I=\int_{-2}^0\frac{1}{x^3}dx+\int_0^3\frac{1}{x^3}dx$$
And then use substitution $x=-t$ we get $$I=\int_2^3\frac1{x^3}dx=\frac5{72}$$
However, If we use limits on both part separately, they both diverge, so the integral diverges too. 
Which explanation is correct?

You are right that the integral is not defined (because of the singularity), so you can't trust what the anti-derivative tells you. It's even worse with $$\int_{-1}^{1}\frac{1}{x^2}dx=-2\tag{1}$$
The area is clearly positive ($+\infty$). If you were to break up $(1)$ you would get $$\int_{-1}^{1}\frac{1}{x^2}dx=2\int_{0}^{1}\frac{1}{x^2}dx\rightarrow\infty$$ So it's not always true that you can break up integrals.
