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

The integral from-infinity to 3 of 1/(x^2+9) is pi/4

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integral from-infinity to 3 of 1/(x^2+9)

\int_{- \infty}^{3} \frac{1}{x ^{2} + 9} d x

\frac{π}{4}

Decimal

0.78539 \dots

Solution steps

\int_{- \infty}^{3} \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_{- \infty}^{3} \frac{1}{x ^{2} + 9} d x = \frac{π}{12} - (- \frac{π}{6})

= \frac{π}{12} - (- \frac{π}{6})

Simplify

= \frac{π}{4}

\frac{\partial}{\partial y} (\frac{6 x}{3 x ^{2} - y ^{3} - 3})

\frac{d}{d x} ((x - 1)^{2} \cdot (x - 2) \cdot (x - 4))

x^{7} - 3 x^{4} + 7 x^{3} + 43, at x = - 2

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

x \to 0 lim (3 - 4 x)

What is the integral from-infinity to 3 of 1/(x^2+9) ?
The integral from-infinity to 3 of 1/(x^2+9) is pi/4