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

The integral from 0 to 1 of 1/(x^2-x) is diverges

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integral from 0 to 1 of 1/(x^2-x)

\int_{0}^{1} \frac{1}{x ^{2} - x} d x

diverges

Solution steps

\int_{0}^{1} \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_{0}^{1} - \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_{0}^{1} \frac{1}{x} d x + \int_{0}^{1} \frac{1}{x - 1} d x

\int_{0}^{1} \frac{1}{x} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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What is the integral from 0 to 1 of 1/(x^2-x) ?
The integral from 0 to 1 of 1/(x^2-x) is diverges