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

The integral from 1 to infinity of 1/(x^2+9) is pi/6-1/3 arctan(1/3)

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

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

\frac{π}{6} - \frac{1}{3} arctan (\frac{1}{3})

Decimal Notation

0.41634 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

\int_{1}^{\infty} \frac{1}{x ^{2} + 9} d x = \frac{π}{6} - \frac{1}{3} arctan (\frac{1}{3})

= \frac{π}{6} - \frac{1}{3} arctan (\frac{1}{3})

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What is the integral from 1 to infinity of 1/(x^2+9) ?
The integral from 1 to infinity of 1/(x^2+9) is pi/6-1/3 arctan(1/3)