What is the integral from 0 to 2 of 1/((x-1)(x-3)) ?

The integral from 0 to 2 of 1/((x-1)(x-3)) is diverges

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integral from 0 to 2 of 1/((x-1)(x-3))

\int_{0}^{2} \frac{1}{( x - 1 ) ( x - 3 )} d x

diverges

Solution steps

\int_{0}^{2} \frac{1}{( x - 1 ) ( x - 3 )} d x

Find undefined (singularity) points:

x = 1

If exist

b, a < b < c, f (b) =

undefined

, \int_{a}^{c} f (x) d x = \int_{a}^{b} f (x) d x + \int_{b}^{c} f (x) d x

= \int_{0}^{1} \frac{1}{( x - 1 ) ( x - 3 )} d x + \int_{1}^{2} \frac{1}{( x - 1 ) ( x - 3 )} d x

\int_{0}^{1} \frac{1}{( x - 1 ) ( x - 3 )} 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 2 of 1/((x-1)(x-3)) ?
The integral from 0 to 2 of 1/((x-1)(x-3)) is diverges