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

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

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

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

diverges

Solution steps

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

Find undefined (singularity) points:

x = \frac{1}{2}

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

\int_{0}^{\frac{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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\int_{0}^{6} \frac{7}{8 x ^{2} + 9 x + 1} d x

\int_{- 1}^{1} \frac{x}{1 + ∣ x ∣} d x

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

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