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

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

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

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

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

Decimal Notation

0.34657 \dots

Solution steps

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

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

Simplify

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

\int 4 \frac{sin ( 2 x )}{sin ( x )} d x

\frac{s - 6}{s + 2}

\int_{0}^{\frac{π}{3}} 81 sin^{2} (x) d x

\int \frac{1}{( x - 2 ) ^{2} ( x ^{2} + 3 ) ^{2}} d x

\frac{d}{d x} (e^{- \frac{x}{2}} cos (\frac{3}{2} x))

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