Evaluate integral of $z/(z^2 + 2z + 5)\,dz$ around the unit circle Evaluate integral
$$\int_c\frac{z}{(z^2 + 2z + 5)}\,dz,$$
where $c$ is the unit circle
I take $u = z^2 +2z$, then $du = 2z+2$, but that not work 
 A: Pretty sure that the answer is zero. By the residue theorem (link), we need to find out how many poles the closed curve (unit circle) winds around. Looking for points where the function is undefined, solve $z^2 + 2z + 5 = 0$ by completing the square; this gives $z = -1 + 2i, -1 - 2i$ and both of these points are clearly outside the unit circle (height $±2i$ in the complex plane). So the curve winds around no poles and it follows that the integral must be zero.
A: Make the substitution $z=e^{i\varphi}, \ \varphi\in [0,2\pi\rangle$, and in the very next step take $t=e^{i\varphi}$. Then you have
$$
\frac{1}{i} \int\limits_0^{2\pi} \! \frac{e^{2i\varphi}}{e^{2i\varphi}+2e^{i\varphi}+5} \, \mathrm{d}\varphi = - \int\limits_1^{1} \! \frac{t}{t^2+2t+5} \, \mathrm{d}t=0.
$$
It's 3 AM so I'm really sorry if I made a mistake. But even if I did make an arithmetic mistake somewhere in my steps, what I've shown would be a good way to go if you want to avoid the Residue theorem or Cauchy's integral formula.
Also, you should check first to see if the roots of the quadratic equation $z^2+2z+5=0$ lie on the unit circle or not, then do this.
A: my answer for the question above 

