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

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

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

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

\frac{π}{6}

Decimal

0.52359 \dots

Solution steps

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

Compute the indefinite integral:

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

Compute the boundaries:

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

= \frac{π}{6} - 0

Simplify

= \frac{π}{6}

\int \frac{1}{( x ^{2} + 2 x )} d x

x^{2} - x^{3} + 4

x \to \infty lim (x^{3} - 1)

\frac{\partial}{\partial x} (x^{2} + y^{2} - x)

y^{^{'}} = - \frac{( 2 x + ln ( y ) ) y}{x}

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