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
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.
my answer for the question above