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

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

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

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

diverges

Solution steps

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

Take the partial fraction of

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

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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