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Since the integral $\int_{0}^{3}\frac{1}{x^3-x^2-x+1}dx$ has a discontinuity inside the interval, we have to split it in two integrals
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$\int_{0}^{1}\frac{1}{x^3-x^2-x+1}dx+\int_{1}^{3}\frac{1}{x^3-x^2-x+1}dx$
Learn how to solve definite integrals problems step by step online. Integrate the function 1/(x^3-x^2-x+1) from 0 to 3. Since the integral \int_{0}^{3}\frac{1}{x^3-x^2-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^3-x^2-x+1}dx results in: \lim_{c\to1}\left(\frac{-1}{2\left(c-1\right)}-\frac{1}{2}\right)+\frac{48}{277}+\lim_{c\to0}\left(\infty \right). Gather the results of all integrals. The integral \int_{1}^{3}\frac{1}{x^3-x^2-x+1}dx results in: \frac{48}{277}+\lim_{c\to1}\left(-\frac{1}{4}+\frac{1}{2\left(c-1\right)}\right)+\lim_{c\to1}\left(-\frac{48}{277}+\frac{1}{4}\ln\left(c-1\right)\right).