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

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

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

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

diverges

Solution steps

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

Compute the indefinite integral:

\int \frac{2}{x ^{2} - 1} d x = - ln ∣ x + 1 ∣ + ln ∣ x - 1 ∣ + C

Compute the boundaries:

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

diverges

= diverges

\int_{0}^{4} 14 (600 + 20 x^{2}) d x

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

\int_{- 6}^{6} 6 - ∣ x ∣ d x

\int_{0}^{2} 4^{3 x} d x

\int_{x}^{1} 6 x d x

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