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

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

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

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

diverges

Solution steps

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

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

diverges

If part of the expression diverges, then entire expression diverges

= diverges

\frac{d}{d x} (3 x^{2} + 4 x + 5)

x \to 0 lim (\frac{x ^{2}}{1 + x ^{2}})

3 (t - π)

x \to \infty lim ((1 - 5 x)^{\frac{5}{x}})

x \to \frac{1}{2} lim (2 x + 7)

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