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

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

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

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

diverges

Solution steps

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

Adjust integral based on undefined pointsThere is an undefined point within the boundaries at

: 1

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

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

diverges

= diverges

\int_{x^{2}}^{2 x} (x^{2} + y^{2}) d y

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

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

\int_{0}^{3} (4^{x} - x^{e}) d x

\int_{- 3}^{3} 4 - x^{2} d x

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