What is the integral from-infinity to 0 of z/(z^4+9) ?

The integral from-infinity to 0 of z/(z^4+9) is -pi/(12)

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integral from-infinity to 0 of z/(z^4+9)

\int_{- \infty}^{0} \frac{z}{z ^{4} + 9} d z

- \frac{π}{12}

Decimal Notation

- 0.26179 \dots

Solution steps

\int_{- \infty}^{0} \frac{z}{z ^{4} + 9} d z

Compute the indefinite integral:

\int \frac{z}{z ^{4} + 9} d z = \frac{1}{6} arctan (\frac{z ^{2}}{3}) + C

Compute the boundaries:

\int_{- \infty}^{0} \frac{z}{z ^{4} + 9} d z = 0 - \frac{π}{12}

= 0 - \frac{π}{12}

Simplify

= - \frac{π}{12}

f (x) = x^{2} - 8

\int_{0}^{1} \frac{1}{x ( x + 2 )} d x

\frac{d}{d x} (\frac{e ^{x}}{x ^{5}})

\frac{d}{d x} ((1 - \frac{x}{7})^{- 7})

\frac{d}{d r} (\frac{1}{r})

What is the integral from-infinity to 0 of z/(z^4+9) ?
The integral from-infinity to 0 of z/(z^4+9) is -pi/(12)