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 Notation

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}

x \to - 5 lim (\frac{1}{x ^{2} - 25})

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

f (x) = ln (x^{3}), at x = e^{5}

f (x) = (x^{3} + 6 x)^{4}, at x = 2

\int_{0}^{2} \frac{2 x}{( x ^{2} + 3 ) ^{3}} d x

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