Can you integrate over a removable discontinuity? $$\int_0^4 x+2\ \mathrm{d}x = 16$$
$$\int_0^4 \frac{x^2-4}{x-2}\,\mathrm{d}x = 16?$$
 A: Let introduce a few notations $\begin{cases} f(x)=\frac{x^2-4}{x-2} \\ \tilde f(x)=x+2\ & \text{its continuous extension}\end{cases}$
$\displaystyle I_1(\varepsilon)=\int_0^{2-\varepsilon}f(x)dx=\int_0^{2-\varepsilon}\tilde f(x)dx=6-4\varepsilon+\varepsilon^2/2\to 6$
$\displaystyle I_2(\varepsilon)=\int_{2+\varepsilon}^4f(x)dx=\int_{2+\varepsilon}^4\tilde f(x)dx=10-4\varepsilon-\varepsilon^2/2\to 10$
Since $\tilde f$ is positive over $[0,4]$, it is clear that both $I_1(\varepsilon)$ and $I_2(\varepsilon)$ are increasing when $\varepsilon$ decreases toward zero.
It is also clear that $I_1(\varepsilon)\le 6$ and $I_2(\varepsilon)\le 10$ for $\varepsilon\ll 1$.
Now, it really depends of the theory of integration you choose. 

From the results above, any method that agrees with the monotone convergence theorem (Beppo-Levi) will automatically agrees with $\displaystyle \int_0^4f(x)dx=\lim\limits_{\varepsilon\to 0}\ I_1(\varepsilon)+I_2(\varepsilon)=16$
This is the case for Lebesgue integral, as well as Riemann-complete integral (also known as Kurzweil-Henstock integral).

For Riemann integral, the integral over $[0,4]$ stays an improper-integral and it is a voluntary choice to extend it to $\lim\limits_{\varepsilon\to 0}\ I_1(\varepsilon)+I_2(\varepsilon)$ by continuity. 
Note that the Lebesgue-like theorem that says that a function is Riemann integrable if its set of discontinuities is of measure zero, is a similar extension of the strict Riemann integration theory, and is also a voluntary choice to extend the value by continuity of the integral.
