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Since the integral $\int_{0}^{3}\frac{1}{x-1}dx$ has a discontinuity inside the interval, we have to split it in two integrals
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$\int_{0}^{1}\frac{1}{x-1}dx+\int_{1}^{3}\frac{1}{x-1}dx$
Learn how to solve definite integrals problems step by step online. Integrate the function 1/(x-1) from 0 to 3. Since the integral \int_{0}^{3}\frac{1}{x-1}dx has a discontinuity inside the interval, we have to split it in two integrals. The integral \int_{0}^{1}\frac{1}{x-1}dx results in: \lim_{c\to0}\left(- \infty \right). The integral \int_{1}^{3}\frac{1}{x-1}dx results in: \lim_{c\to1}\left(\ln\left(2\right)-\ln\left(c-1\right)\right). Gather the results of all integrals.