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

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

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

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

diverges

Solution steps

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

Take the partial fraction of

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

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

\int_{0}^{\frac{1}{2}} \frac{1}{x - 1} d x = - ln (2)

\int_{0}^{\frac{1}{2}} \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 1/2 of 1/((1-x)x^2) ?
The integral from 0 to 1/2 of 1/((1-x)x^2) is diverges