What is the integral from 1 to 2 of 1/(x^2-x) ?

The integral from 1 to 2 of 1/(x^2-x) is diverges

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integral from 1 to 2 of 1/(x^2-x)

\int_{1}^{2} \frac{1}{x ^{2} - x} d x

diverges

Solution steps

\int_{1}^{2} \frac{1}{x ^{2} - x} d x

Take the partial fraction of

\frac{1}{x ^{2} - x} : - \frac{1}{x} + \frac{1}{x - 1}

= \int_{1}^{2} - \frac{1}{x} + \frac{1}{x - 1} d x

Apply the Sum Rule:

\int f (x) \pm g (x) d x = \int f (x) d x \pm \int g (x) d x

= - \int_{1}^{2} \frac{1}{x} d x + \int_{1}^{2} \frac{1}{x - 1} d x

\int_{1}^{2} \frac{1}{x} d x = ln (2)

\int_{1}^{2} \frac{1}{x - 1} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 1 to 2 of 1/(x^2-x) ?
The integral from 1 to 2 of 1/(x^2-x) is diverges