What is the integral from 4 to infinity of 1/(x^2-9) ?

The integral from 4 to infinity of 1/(x^2-9) is (ln(7))/6

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integral from 4 to infinity of 1/(x^2-9)

\int_{4}^{\infty} \frac{1}{x ^{2} - 9} d x

\frac{ln ( 7 )}{6}

Decimal

0.32431 \dots

Solution steps

\int_{4}^{\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_{4}^{\infty} \frac{1}{x ^{2} - 9} d x = 0 - (- \frac{ln ( \frac{7}{3} ) + ln ( 3 )}{6})

= 0 - (- \frac{ln ( \frac{7}{3} ) + ln ( 3 )}{6})

Simplify

= \frac{ln ( 7 )}{6}

\int_{- 7}^{7} x^{2} d x

\int_{0}^{2 π} - 7 sin^{2} (t) d t

\int_{0}^{2 π} (cos (θ))^{2} d θ

\int_{0}^{1} 4 e^{x} d x

\int_{12}^{x} \frac{1}{t} d t

What is the integral from 4 to infinity of 1/(x^2-9) ?
The integral from 4 to infinity of 1/(x^2-9) is (ln(7))/6