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Since the integral $\int_{-2}^{1}\frac{1}{x^2}dx$ has a discontinuity inside the interval, we have to split it in two integrals

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$\int_{-2}^{0}\frac{1}{x^2}dx+\int_{0}^{1}\frac{1}{x^2}dx$

Learn how to solve limits to infinity problems step by step online. Integrate the function 1/(x^2) from -2 to 1. Since the integral \int_{-2}^{1}\frac{1}{x^2}dx has a discontinuity inside the interval, we have to split it in two integrals. The integral \int_{-2}^{0}\frac{1}{x^2}dx results in: \lim_{c\to0}\left(\frac{1}{-c}-\frac{1}{2}\right). The integral \int_{0}^{1}\frac{1}{x^2}dx results in: \lim_{c\to0}\left(-1+\frac{1}{c}\right). Gather the results of all integrals.

** Final answer to the problem

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