# What is the integral of #int (x-2)/(x-1)# from 0 to 2?

We have:

And, slightly trickier:

For the previous integral, notice that the derivative of the denominator is present in the numerator. This means that we have a natural logarithm integral present.

Combining these, we want to evaluate

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To solve the integral of (\int_{0}^{2} \frac{x-2}{x-1} , dx), you need to decompose the integrand using partial fraction decomposition. The integral will then become (\int_{0}^{2} (1 - \frac{1}{x-1}) , dx). Integrating term by term, you get (x - \ln|x-1|). Evaluating this expression from 0 to 2, you substitute 2 and 0 into the expression and subtract the result of plugging in 0 from the result of plugging in 2. This yields (2 - \ln|1| - (0 - \ln|-1|) = 2 - \ln|1| + \ln|1| = 2). Therefore, the integral of (\int_{0}^{2} \frac{x-2}{x-1} , dx) from 0 to 2 is equal to 2.

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