How do you integrate this complex function The integration is a closed integral/circulation (I couldn't find out the mathjax for closed integral sign !)
$$ \int_c \frac{-3z+4}{z^2+4z+5}\, dz, \space \space \space \space \space \text {where $\Bbb c$ is the circle } |z| = 1
$$
My question is
 - In order to evaluate this integral, what/which concept(s) do I need to go through?
 - Is there a general approach to solve this kind of questions like  


*

*Do a partial fraction

*Do Substitution/ Integrate by parts/ apply formula ...


I have done some elementary calculus and complex analysis but this is something I don't find familiar.
 A: The only poles of this rational function are at $-2\pm i$.  Those are OUTSIDE the circle.  So what's the integral around a curve of any function that's holomorphic on and inside the curve?
A: You're forgetting to take into consideration the curve $C$! Let's parametrize the curve. We are looking at $|z|=1$. This is the unit circle. So we will parametrize in the 'typical' fashion, $z=e^{it}$ for $t \in [0,2\pi]$. Then $dz=ie^{it}dt$ then what we have is the integral
$$
\int_C \frac{-3z+4}{z^2+4z+5}\;dz=\int_0^{2\pi} \frac{-3e^{it}+4}{e^{2it}+4e^{it}+5}\;dt
$$
Of course, this really isn't much different than the integral we started with. Rewrite the integral as 
$$
\int_0^{2\pi} \frac{10}{e^{2it}+4e^{it}+5}-\frac{3(2e^{it}+4)}{2(e^{2it}+4e^{it}+5)}\;dt
$$
The first term can be integrated by completing the square, making a substitution, then using inverse functions (a standard technique). The second term can be solved using a simple $u$-substitution.
This is the long way of doing this problem. Since you hinted at only an elementary approach to the problem, this method only requires basic Calculus techniques after the initial parametrization. However, the fast way of doing this problem is to use the Residue Theorem. The denominator is only zero when $z^2+4z+5=0$, that is when $z=-2\pm i$. These "correspond" to the points $(-2,\pm 1)$ in $\mathbb{R}^2$. However, neither of these are inside our closed curve $C$-the unit circle. Therefore, the Residue Theorem says.........

 The integral is $0$. All that work for nothing!

