What is wrong with the following calculation?

​$$\int_{-1}^3 \frac{1}{x^2} d x=-\frac{4}{3}$$​​

Correct option is B

​$$\begin{aligned}&\text{For improper integrals with discontinuities, we must split the integral at the point of discontinuity and take limits:} \\&\int_{-1}^{3} \frac{1}{x^2} \, dx = \int_{-1}^{0} \frac{1}{x^2} \, dx + \int_{0}^{3} \frac{1}{x^2} \, dx \\&\text{Evaluate each part separately using limits:} \\&\text{1. For } \int_{-1}^{0} \frac{1}{x^2} \, dx: \\&\lim_{a \to 0^-} \int_{-1}^{a} \frac{1}{x^2} \, dx = \lim_{a \to 0^-} \left[ -\frac{1}{x} \right]_{-1}^{a} = \lim_{a \to 0^-} \left( -\frac{1}{a} + 1 \right) = +\infty \\&\text{2. For } \int_{0}^{3} \frac{1}{x^2} \, dx: \\&\lim_{b \to 0^+} \int_{b}^{3} \frac{1}{x^2} \, dx = \lim_{b \to 0^+} \left[ -\frac{1}{x} \right]_{b}^{3} = \lim_{b \to 0^+} \left( -\frac{1}{3} + \frac{1}{b} \right) = +\infty \\&\text{Both parts diverge to } +\infty, \text{ so the original integral \textbf{does not converge}.} \\\\&\textbf{Conclusion} \\&\text{The error in the original calculation is that it ignores the discontinuity at } x = 0. \\&\text{The integral is improper and diverges (i.e., it does not exist as a finite value).} \\\\&\text{The calculation is incorrect because the integral } \int_{-1}^{3} \frac{1}{x^2} \, dx \text{ is improper (due to the discontinuity at } x = 0\text{) and diverges.} \\&\text{The correct evaluation shows that the integral does not converge to } -\frac{4}{3} \text{ or any finite value.} \\\\&{\text{The integral diverges because } \frac{1}{x^2} \text{ is undefined at } x = 0 \text{ and the limits approach } +\infty.}\end{aligned}$$​​