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

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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 Notation

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}

\frac{d}{d x} (e^{x} cosh (x))

x \to 0 - lim (\frac{x}{8 x ^{4} + 2 x})

x \to 3 - lim ((x - 2)^{3})

\frac{0.9}{0.1 s ^{2} + s + 0.9}

\int \frac{5 + x}{x} d x

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