What is the integral from 4 to infinity of 1/(x^2-1) ?

The integral from 4 to infinity of 1/(x^2-1) is -(ln(3)-ln(5))/2

English

Español

Português

Français

Deutsch

Italiano

Русский

中文(简体)

한국어

日本語

Tiếng Việt

עברית

العربية

integral from 4 to infinity of 1/(x^2-1)

\int_{4}^{\infty} \frac{1}{x ^{2} - 1} d x

- \frac{ln ( 3 ) - ln ( 5 )}{2}

Decimal

0.25541 \dots

Solution steps

\int_{4}^{\infty} \frac{1}{x ^{2} - 1} d x

Compute the indefinite integral:

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

Compute the boundaries:

\int_{4}^{\infty} \frac{1}{x ^{2} - 1} d x = 0 - \frac{- ln ( 5 ) + ln ( 3 )}{2}

= 0 - \frac{- ln ( 5 ) + ln ( 3 )}{2}

Simplify

= - \frac{ln ( 3 ) - ln ( 5 )}{2}

What is the integral from 4 to infinity of 1/(x^2-1) ?
The integral from 4 to infinity of 1/(x^2-1) is -(ln(3)-ln(5))/2