Does $\int_1^2 \frac{\ln(x)}{x-1} dx$ converge and what test is used? $$\int_1^2 \frac{\ln(x)}{x-1} dx$$
How does one determine convergence of this? I am not interested in the value of it. I tried comparing to $1/(x-1)$ but the integral related to that diverges, and I know that above converges (via mathematica) so I need something that converges to compare with, but what?

It is easier to handle if we look at
$$I = \int_1^2 \dfrac{\ln(x)}{x-1}dx = \int_0^1 \dfrac{\ln(1+t)}{t}dt$$
Note that for $t \in (0,1)$, we have $\dfrac{\ln(1+t)}t \in (0,1)$. Hence, clearly, $I$ exists and the value is in fact between $0$ and $1$. To evaluate the integrate note that
$$\ln(1+t) = \sum_{k=1}^{\infty}(-1)^{k+1} \dfrac{t^k}k$$
Hence, we have
\begin{align}
I & = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}k \int_0^1 t^{k-1}dt = \sum_{k=1}^{\infty} \dfrac{(-1)^k}k \cdot \left.\dfrac{t^k}k \right \vert_0^1 = \sum_{k=1}^{\infty} \dfrac{(-1)^{k+1}}{k^2} = \dfrac1{1^2} - \dfrac1{2^2} + \dfrac1{3^2}-\dfrac1{4^2} \pm \cdots\\
& = \left(\dfrac1{1^2} + \dfrac1{2^2} + \dfrac1{3^2} + \dfrac1{4^2} + \cdots \right) - 2\left(\dfrac1{2^2} + \dfrac1{4^2} + \cdots\right) = \zeta(2) - \dfrac12 \zeta(2) = \dfrac{\zeta(2)}2 = \dfrac{\pi^2}{12}
\end{align}
