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Since the integral $\int_{0}^{3}\frac{6}{x^2-5x+4}dx$ has a discontinuity inside the interval, we have to split it in two integrals
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$\int_{0}^{1}\frac{6}{x^2-5x+4}dx+\int_{1}^{3}\frac{6}{x^2-5x+4}dx$
Learn how to solve problems step by step online. Integrate the function 6/(x^2-5x+4) from 0 to 3. Since the integral \int_{0}^{3}\frac{6}{x^2-5x+4}dx has a discontinuity inside the interval, we have to split it in two integrals. The integral \int_{0}^{1}\frac{6}{x^2-5x+4}dx results in: undefined. The integral \int_{1}^{3}\frac{6}{x^2-5x+4}dx results in: undefined. Gather the results of all integrals.