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Since the integral $\int_{-2}^{2}\frac{1}{x\sqrt{16-x^2}}dx$ has a discontinuity inside the interval, we have to split it in two integrals
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$\int_{-2}^{0}\frac{1}{x\sqrt{16-x^2}}dx+\int_{0}^{2}\frac{1}{x\sqrt{16-x^2}}dx$
Learn how to solve problems step by step online. Integrate the function 1/(x(16-x^2)^1/2) from -2 to 2. Since the integral \int_{-2}^{2}\frac{1}{x\sqrt{16-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\sqrt{16-x^2}}dx results in: \lim_{c\to-2}\left(- \infty \right). The integral \int_{0}^{2}\frac{1}{x\sqrt{16-x^2}}dx results in: \lim_{c\to0}\left(-0.329239+\frac{1}{4}\ln\left(\frac{4}{c}+\frac{\sqrt{16-c^2}}{c}\right)\right). Gather the results of all integrals.