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?

On account of the singularity at $x=0$, the integral you give is undefined. The only way you could get close is with:
$$\int_{-2}^{3}{[x^{-3}dx]}\approx\int_{-2}^{-\epsilon}{[x^{-3}dx]}+\int_{\epsilon}^{3}{[x^{-3}dx]}$$
With $\epsilon$ as a very small positive constant.
