What is the integral from 1 to 2 of 1/((x-1)(x-3)) ?

The integral from 1 to 2 of 1/((x-1)(x-3)) is diverges

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integral from 1 to 2 of 1/((x-1)(x-3))

\int_{1}^{2} \frac{1}{( x - 1 ) ( x - 3 )} d x

diverges

Solution steps

\int_{1}^{2} \frac{1}{( x - 1 ) ( x - 3 )} d x

Take the partial fraction of

\frac{1}{( x - 1 ) ( x - 3 )} : - \frac{1}{2 ( x - 1 )} + \frac{1}{2 ( x - 3 )}

= \int_{1}^{2} - \frac{1}{2 ( x - 1 )} + \frac{1}{2 ( x - 3 )} 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}{2 ( x - 1 )} d x + \int_{1}^{2} \frac{1}{2 ( x - 3 )} d x

\int_{1}^{2} \frac{1}{2 ( 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)(x-3)) ?
The integral from 1 to 2 of 1/((x-1)(x-3)) is diverges