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Since the integral $\int_{-2}^{2}\frac{1}{x^2-1}dx$ has a discontinuity inside the interval, we have to split it in two integrals
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$\int_{-2}^{-1}\frac{1}{x^2-1}dx+\int_{-1}^{2}\frac{1}{x^2-1}dx$
Learn how to solve definite integrals problems step by step online. Integrate the function 1/(x^2-1) from -2 to 2. Since the integral \int_{-2}^{2}\frac{1}{x^2-1}dx has a discontinuity inside the interval, we have to split it in two integrals. Since the integral \int_{-1}^{2}\frac{1}{x^2-1}dx has a discontinuity inside the interval, we have to split it in two integrals. The integral \int_{-2}^{-1}\frac{1}{x^2-1}dx results in: undefined. The integral \int_{-1}^{1}\frac{1}{x^2-1}dx results in: \lim_{c\to-1}\left(-\frac{96}{277}+\frac{1}{2}\ln\left(c+1\right)\right)+\lim_{c\to-1}\left(- \infty \right).