This is an example from my textbook
I'm just thinking why do we have to break the very right-hand side of the integral in red into two parts again, can we make it like $$\lim_{R_{1} \to{1^{+}},R_{2} \to \infty}\int_{R_{1}}^{R_{2}}\frac{1}{(x-1)^{2}}dx$$
To some extent, it's because when computing limits like
$$\lim_{R_{1} \to{1^{+}},R_{2} \to \infty}\int_{R_{1}}^{R_{2}}\frac{1}{(x-1)^{2}}dx$$
there can be ambiguity over how to compute the limit, e.g. do you take $R_1 \to1$, then $R_2\to\infty$, or do we go in the opposite order, or do we compute them simultaneously (e.g. let $R_1=1+1/t, R_2=t$, and take $t\to\infty$)?
By splitting the integral up, we avoid this problem, as it lets us treat the two limits independently, which is generally simpler.
For an example of why this is important, consider trying to calculate the (divergent) integral $\int_0^\infty\left(2x-\frac{2}{x^3}\right)\,dx$ by the following approaches (as $u\to\infty$):
$$\int_{1/u}^u\left(2x-\frac{2}{x^3}\right)\,dx = 0$$
$$\int_{1/2u}^u\left(2x-\frac{2}{x^3}\right)\,dx = -3u^2+\frac{3}{4u^2}\to -\infty$$
$$\int_{2/u}^u\left(2x-\frac{2}{x^3}\right)\,dx = \frac{3u^2}{4}-\frac{3}{u^2} \to \infty$$
So, any attempts to compute both limits simultaneously requires a bit of care. It's thus often easiest to isolate each limit in a way that allows you to treat them independently.