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

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

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

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

diverges

Solution steps

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

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

: 1

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

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

diverges

= diverges

x^{2} 8 x - 9

(1 + x)^{(- 4)}, 0

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

\int_{- 125}^{64} \frac{1}{x ^{\frac{1}{3}}} d x

\frac{\partial}{\partial y} (x y z^{2})

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