What is the integral from 0 to infinity of 1/(x^2+9) ?

The integral from 0 to infinity of 1/(x^2+9) is pi/6

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integral from 0 to infinity of 1/(x^2+9)

\int_{0}^{\infty} \frac{1}{x ^{2} + 9} d x

\frac{π}{6}

Decimal

0.52359 \dots

Solution steps

\int_{0}^{\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_{0}^{\infty} \frac{1}{x ^{2} + 9} d x = \frac{π}{6} - 0

= \frac{π}{6} - 0

Simplify

= \frac{π}{6}

\int \frac{x - 1}{x} d x

d er i v a t i v e sin (\frac{π x}{3})

d er i v a t i v e f (x) = \frac{3}{2 x - 4}

d er i v a t i v e e^{2 s i n (3 v)}

\frac{d}{d x} (x y + ln (\frac{x}{y}))

What is the integral from 0 to infinity of 1/(x^2+9) ?
The integral from 0 to infinity of 1/(x^2+9) is pi/6