What is the integral from 8 to infinity of 1/(x^2-9) ?

The integral from 8 to infinity of 1/(x^2-9) is (ln(11/3)-ln(5/3))/6

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integral from 8 to infinity of 1/(x^2-9)

\int_{8}^{\infty} \frac{1}{x ^{2} - 9} d x

\frac{ln ( \frac{11}{3} ) - ln ( \frac{5}{3} )}{6}

Decimal Notation

0.13140 \dots

Solution steps

\int_{8}^{\infty} \frac{1}{x ^{2} - 9} d x

Compute the indefinite integral:

\int \frac{1}{x ^{2} - 9} d x = - \frac{1}{6} (ln \frac{x}{3} + 1 - ln \frac{x}{3} - 1) + C

Compute the boundaries:

\int_{8}^{\infty} \frac{1}{x ^{2} - 9} d x = 0 - (- \frac{ln ( \frac{11}{3} ) - ln ( \frac{5}{3} )}{6})

= 0 - (- \frac{ln ( \frac{11}{3} ) - ln ( \frac{5}{3} )}{6})

Simplify

= \frac{ln ( \frac{11}{3} ) - ln ( \frac{5}{3} )}{6}

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What is the integral from 8 to infinity of 1/(x^2-9) ?
The integral from 8 to infinity of 1/(x^2-9) is (ln(11/3)-ln(5/3))/6