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

The integral from 2 to infinity of 1/(1-x^2) is -1/2 ln(3)

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

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

- \frac{1}{2} ln (3)

Decimal Notation

- 0.54930 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

\int_{2}^{\infty} \frac{1}{1 - x ^{2}} d x = 0 - \frac{1}{2} ln (3)

= 0 - \frac{1}{2} ln (3)

Simplify

= - \frac{1}{2} ln (3)

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What is the integral from 2 to infinity of 1/(1-x^2) ?
The integral from 2 to infinity of 1/(1-x^2) is -1/2 ln(3)