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

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

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

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

diverges

Solution steps

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

Take the partial fraction of

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

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

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