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

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

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

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

\frac{π}{6}

Decimal

0.52359 \dots

Solution steps

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

= 0 - (- \frac{π}{6})

Simplify

= \frac{π}{6}

(- 1, 4), (5, - 1)

y^{^{'}} = ry ln (\frac{k}{y})

x \to + 5 + lim (x - 5)

\frac{\partial}{\partial x} (ln (6 x^{2} + y^{2} + 4))

y = x^{2} - 4 x + 4, (3, 1)

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