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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

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

\int_{0}^{2} \frac{x ^{6}}{1 + x ^{2}} d x

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

\int_{0}^{2} 2 x ln (1 + x) d x

\int_{1}^{2} \frac{x ^{2}}{2} d x

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