## Calculus: Early Transcendentals 8th Edition

$-\displaystyle \frac{\pi}{8}.\quad$ (converges)
$\displaystyle \int\frac{z}{z^{4}+4}dz=\left[\begin{array}{l} u=z^{2}\\ du=2zdz \end{array}\right]=\frac{1}{2}\int\frac{du}{u^{2}+2^{2}}du$ Table of integrals: 17. $\displaystyle \int\frac{du}{a^{2}+u^{2}}=\frac{1}{a}\tan^{-1}\frac{u}{a}+C$ The integral is type I improper integral, so $\displaystyle \int_{-\infty}^{0}\frac{z}{z^{4}+4}dz=\lim_{t\rightarrow-\infty}\int_{t}^{0}\frac{z}{z^{4}+4}dz=\lim_{t\rightarrow-\infty}\frac{1}{2}\left[\frac{1}{2}\tan^{-1}\left(\frac{z^{2}}{2}\right)\right]_{t}^{0}\quad$ $=\displaystyle \lim_{t\rightarrow-\infty}\left[0-\frac{1}{4}\tan^{-1}\left(\frac{t^{2}}{2}\right)\right]$ $=-\displaystyle \frac{1}{4}\left(\frac{\pi}{2}\right)$ $=-\displaystyle \frac{\pi}{8}.\quad$ (converges)