$\int_{-\infty}^{\infty} \frac{1}{(z^{4}-1)}\,dz$ $\int_{-\infty}^{\infty} \frac{1}{(z^{4}-1)}\,dz$
$\text{then} \ z=i,-i,-1,1$ are the value. and Do I have to solve all of these residue??
I cannot set which of that is value for residue.
 A: You can either evaluate it with Residues or with the Cauchy principal Value, which may be worth to be written for once. 
$$\text{P.V.} \int_{-\infty}^{+\infty}\frac{1}{z^4-1}\ \text{d}z = \lim_{R\to \infty}\int_{-R}^{+R}\frac{1}{z^4-1}\ \text{d}z$$
The integral is quite easy to evaluate. You can do it!
The result is:
$$\lim_{R\to \infty}\left[-\frac{1}{2}\arctan(z) + \frac{1}{4}\log\left(\frac{1-z}{1+z}\right)\right]_{-R}^{+R}$$
$$\lim_{R\to \infty}\left[-\frac{1}{2}\left(\arctan(R) - \arctan(-R)\right) + \frac{1}{4}\left[\log\left(\frac{1-R}{1+R}\right) - \log\left(\frac{1-R}{1+R}\right)\right]\right]$$
As $R$ goes to infinity, the arctangent is well defined. The logarithmic team, instead, goes to zero as you may see by computing the limit.
Hence
$$\text{P.V.} \int_{-\infty}^{+\infty}\frac{1}{z^4-1}\ \text{d}z = -\frac{1}{2}\left(\frac{\pi}{2} + \frac{\pi}{2}\right) = -\frac{\pi}{2}$$
As wanted.
A: It seems some people misunderstand the principal value (p.v.).  In this case, the principal value integral is
$$
\lim_{a \searrow 0} \left(\int_{-\infty}^{-1-a}\frac{1}{z^4-1}
+\int_{-1+a}^{1-a}\frac{1}{z^4-1}
+\int_{1+a}^{+\infty}\frac{1}{z^4-1}\right)
$$
This is, indeed, $-\pi/2$ as Alan obtained by glossing over the difficulties.  
In Alan's calculation, the fact that
$$
\log\left(\frac{1-z}{1+z}\right)
$$
goes to zero at $\pm \infty$ is irrelevant, since it is discontinuous at $1$ and $-1$.
