Integrating $x/(x-2)$ from $0$ to $5$ How would one go about integrating the following?
$$\int_0^5 \frac{x}{x-2} dx$$
It seems like you need to use long division, split it up into two integrals, and the use limits. I'm not quite sure about the limits part.

Yes, exactly, you do want to use "long division"...
Note, dividing the numerator by the denominator gives you:
$$\int_0^5 {x\over{x-2}} \mathrm{d}x = \int_0^5 \left(1 + \frac 2{x-2}\right) \mathrm{d}x$$
Now simply split the integral into the sum of two integrals:
$$\int_0^5 \left(1 + \frac 2{x-2}\right) \mathrm{d}x \quad= \quad\int_0^5 \,\mathrm{d}x \;\; + \;\; 2\int_0^5 \frac 1{x-2} \,\mathrm{d}x$$
The problem, of course, is what happens with the limits of integration in the second integral: If $u = x-2$ then the limits of integration become $\big|_{-2}^3$, and we will see that the integral does not converge - there is a discontinuity at $u = 0$, and indeed, a vertical asymptote at $x = -2$, hence the limit of the integral - evaluated as $u \to 0$ does not exist, so the integral does not converge. And so the sum of the integrals will not converge.
