Question: I just can't get past the $x-1$. The $2$ I know how to get rid of.

$$\lim\_{x\to +∞}\frac {\ln\left((x-1)^2\right)}{x} $$

Answer: 
$$\lim\_{x\to \infty} \frac{\ln(x-1)^2}{x}$$

It is not hard to show $\ln x< x$

$\ln(x-1)^2=4\ln\sqrt{x-1}\leq 4\sqrt{x-1}\leq 4\sqrt x$

So the limit is smaller than

$$\lim\_{x\to \infty} \frac{4\sqrt{x}}{x}=\lim\_{x\to \infty} \frac 4{\sqrt x}=0$$

Now the following is true because square root must be non negative

$-4\sqrt x\leq ln(x-1)^2$

So the limit is bounded by $0$
Use *equivalents*:

$(x-1)^2\sim\_\infty x^2$ and none of them approach $1$ in a neighbourhood of $+\infty$, hence $\;\ln(x-1)^2\sim\_\infty \ln x^2=2\ln x\;$ and finally
$$\frac{\ln(x-1)^2}{x}\sim\_\infty2\,\frac{\ln x}x \to 0.$$

*Note:* In general equivalence of functions is not compatible with composition on the left with another function.