What is the integral from 0 to 2 of 1/(x^2-4x+3) ?

The integral from 0 to 2 of 1/(x^2-4x+3) is diverges

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integral from 0 to 2 of 1/(x^2-4x+3)

\int_{0}^{2} \frac{1}{x ^{2} - 4 x + 3} d x

diverges

Solution steps

\int_{0}^{2} \frac{1}{x ^{2} - 4 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 ^{2} - 4 x + 3} d x + \int_{1}^{2} \frac{1}{x ^{2} - 4 x + 3} d x

\int_{0}^{1} \frac{1}{x ^{2} - 4 x + 3} d x =

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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\int \frac{1}{16 x ^{2} - 81} d x

What is the integral from 0 to 2 of 1/(x^2-4x+3) ?
The integral from 0 to 2 of 1/(x^2-4x+3) is diverges