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

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

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

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

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

Decimal

0.54930 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

\int_{2}^{\infty} \frac{1}{y ^{2} - 1} d y = 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/(y^2-1) ?
The integral from 2 to infinity of 1/(y^2-1) is 1/2 ln(3)