Why is it not possible to integrate over a singularity? I came across a question that states that the following integral does not exist
$$\displaystyle \int_{-4}^{4} \frac{1}{x+2} \,\mathrm dx$$
I have seen similar questions on here where people have said that they can be integrated and have shown this. I am confused as to whether it can be or not. 
 A: It depends on what type of integral you are considering. For example the integral from $-4$ to $-2-\epsilon$ plus the integral from $-2+\epsilon $ to $0$ vanishes for every $\epsilon >0$ so the given integral is $\int_0^{4} \frac  1{x+2} dx=ln(6)-ln(2)=\ln(3)$ in a limiting sense. But the integral from $-2$ to $4$ is $\infty$ by direct calculation , so in the usual sense the integral does not exist. 
I am thankful to Raymond Manzoni for his help in correcting a mistake. 
A: Strictly something like this cannot be integrated, as this singularity causes several problems. If we were to just calculate the value using normal calculus and ignore the presence of this we would find the Cauchy principle value, however this effectively assumes that the asymptote upwards is equal in magnitude to the one downwards, and so they 'cancel'. However, in most areas of maths you cannot assume "all infinities are equal" and so this would be seen as divergent. 
A: The Riemann integral $\displaystyle \int_{-4}^{4} \frac{1}{x+2} \,\mathrm dx$ as no defined value since the function $\frac{1}{x+2}$ is not finite at $x=-2$ and this point is in the range of integration.
But the integral has a different meaning in sens of Cauchy Principal Value :
$$PV\displaystyle \int_{-4}^{4} \frac{1}{x+2} \,\mathrm dx = \lim_{\epsilon\to 0}\left(\displaystyle \int_{-4}^{-2-\epsilon} \frac{1}{x+2} \,\mathrm dx +\displaystyle \int_{-2+\epsilon}^{4} \frac{1}{x+2} \,\mathrm dx\right)= \ln(3)$$
http://mathworld.wolfram.com/CauchyPrincipalValue.html
So take carre that the answer is not the same if the question refers to the value of a definite Riemann integral and an integral in sens of Cauchy Principal value which is denoted with $PV$ in front of the integral.
