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

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

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integral from 3 to infinity of 1/(x^2-4)

\int_{3}^{\infty} \frac{1}{x ^{2} - 4} d x

\frac{ln ( 5 )}{4}

Decimal Notation

0.40235 \dots

Solution steps

\int_{3}^{\infty} \frac{1}{x ^{2} - 4} d x

Compute the indefinite integral:

\int \frac{1}{x ^{2} - 4} d x = - \frac{1}{4} (ln \frac{x}{2} + 1 - ln \frac{x}{2} - 1) + C

Compute the boundaries:

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

= 0 - (- \frac{ln ( \frac{5}{2} ) + ln ( 2 )}{4})

Simplify

= \frac{ln ( 5 )}{4}

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