Consider a Cauchy principal value integral over the real line $$ I = \operatorname{PV} \int_{-\infty}^{\infty} dx f(x) \tag{1} $$ where $f(x)$ is a function which is continuous but for finitely many poles $x_n$, all of which are simple, and $f(x)$ decays sufficiently fast as $x \to \pm \infty$ for integrals $\int_{- \infty}^a dx f(x)$, $\int_{b}^\infty dx f(x)$ to converge for $a < \min_n x_n$, $\max_n x_n < b$.
Consider the proposition that the principal value $I$ is equal to the convergent integral $$ I = \int_{-\infty}^{\infty} d x g(x) \tag2 $$ where $g(x)$ is a function with no poles given by $$ g(x) = f(x) - \sum_n\frac{\lim_{y\to x_n} f(y)(y-x_n)}{x-x_n}. $$ Is this proposition true?
For example, for $f(x) = ( \cosh x - \cosh 2)^{-1}$ we have $g(x) = ( \cosh x - \cosh 2)^{-1} - \operatorname{csch} 2 ((x/2)^2 - 1)^{-1}$ and thus obtain $I = 8 e^2 / (1-e^4)$ by evaluating either (1) or (2).
The argument for the proposition is that I am free to add functions of the form $a(x-b)^{-1}$ to the integrand of (1) without altering the principal value of the integral, and can remove all the poles in this manner.
Follow up: can a similar trick be used for higher order poles?
The "principal value" at a pole of order $1$ works like this, since the two sides "cancel" when you approach at the same rate.
$$ \int_{x_n-1}^{x_n-\epsilon} \frac{dx}{x-x_n} + \int_{x_n+\epsilon}^{x_n+1}\frac{dx}{x-x_n} = 0 . $$
Another point. $$ g(x) = f(x) - \sum_n\frac{\lim_{y\to x_n} f(y)(y-x_n)}{x-x_n}. $$ has no poles. But its integral may no longer converge at $\pm \infty$.
Example $$ f(x) = \frac{1}{x^3-x} $$ has simple pole at $x=0$, and behaves like $1/|x|^2$ as $x\to\infty$. But $$ g(x) = \frac{1}{x^3-x} - \frac{1}{x} = \frac{-x}{x^2+1} $$ behaves like $1/|x|$ as $|x| \to \infty$, so the integral diverges. For a PV you will have to use $$ \lim_{N\to\infty} \int_{-N}^{N} \frac{-x\;dx}{x^2+1} . $$
But an integral with a pole of order $2$ diverges even if you attempt a principle value. Also (usually) poles of order higher than $2$.