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

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

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

\int_{3}^{\infty} \frac{2}{x ^{2} - 1} d x

ln (2)

Decimal

0.69314 \dots

Solution steps

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

= 0 - (- ln (2))

Simplify

= ln (2)

ln (1 + z)

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\int_{1}^{e^{3}} \frac{ln ( x )}{x ^{2}} d x

y^{^{'}} + y = e^{t}

\int \frac{x ^{2} - 9}{x ^{3}} d x

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