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

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

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

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