Finding Cauchy principal value for: $ \int_1^\infty \frac{ a x^2 + c }{x^4 - b x^2 - c} \mathrm{d}x $ I need to solve the integral
$ \displaystyle \mathcal{P} \int_1^\infty \frac{ a x^2 + c }{x^4 - b x^2 - c} \mathrm{d}x $,
where $\mathcal{P}$ is the Cauchy principal value, $ - 1 \leq c \leq 1$ and $a, b$ are both real, but can be arbitrarily large, positive or negative.
I'm not sure, whether this integral is solveable, but any hints or ideas are very welcome.
 A: $$\frac A{x^2-r_0}+\frac B{x^2-r_1}=\frac{(A+B)x^2-(r_1A+r_0B)}{x^4-bx^2-c},$$
$$A+B=a\\r_1A+r_0B=-c,$$
$$A=\frac{r_0a+c}{r_0-r_1},\\
B=\frac{r_1a+c}{r_1-r_0}.$$
The complete discussion must take into account the cases of complex and/or negative roots, but the integration is straightforward.
$$\int\frac{dx}{x^2-r}=\frac1{2\sqrt r}\ln\frac{x-\sqrt r}{x+\sqrt r}+C.$$
The terms at infinity will vanish ($\ln1$), as the integrand $O(\frac1{x^2})$ is convergent. Single or double real roots $>1$ will cause divergence.
A: I actually ended up using a different solution, I found more direct and intuitively.
It is completely equivalent with @Yves method, but I just state it for completeness.
$$ \mathcal{P} \int_{1}^{\infty} \mathrm{d} x
    \frac{a  x^{2} + c }
     {
      x^{4} - b  x^{2} - c
     }
  =
   \mathcal{P} \int_{1}^{\infty} \mathrm{d} x
   \frac{a  x^{2} + c }{ (x^{2} - d)^{2} - g^{2}}
$$
$$
  = \frac{1}{2g}
   \Bigg\{
    \mathcal{P} \int_{1}^{\infty}
     a x^{2}
    \left[
     \frac{1}{ x^{2} - d - g}
    -
     \frac{1}{ x^{2} - d - g}
    \right]
    \mathrm{d} x
$$
$$
   \qquad \quad +
    \mathcal{P} \int_{1}^{\infty}
    c
    \left[
     \frac{1}{ x^{2} - d - g}
    -
     \frac{1}{ x^{2} - d - g}
    \right]
     \mathrm{d} x
   \Bigg\}
$$
with
$$ d = - \frac{b}{2}, \qquad g = \sqrt{c + d^2}. $$
Then dividing each fraction in two once again, we end with two integrals to solve
$$
 \mathcal{P} \int_{1}^{\infty} \mathrm{d} x \frac{x}{x \pm \sqrt{d \pm g}}
=
  \lim_{\epsilon \rightarrow \infty} \left[ x \mp \sqrt{r} \ln( x \pm \sqrt{r} ) \right]_{x = 1}^{x = \epsilon},
$$
$$
 \mathcal{P} \int_{1}^{\infty} \mathrm{d} x \frac{1}{x \pm \sqrt{d \pm g}} 
=
  \lim_{\epsilon \rightarrow \infty} \left[ \sqrt{r} \ln( x \pm \sqrt{r} ) \right]_{x = 1}^{x = \epsilon}.
$$
Then, putting it all together, we end up with
$$
\frac{1}{8 g} \left[
     \frac{2 a (d + g) + c}{\sqrt{d + g}} \ln \left( \frac{1 + \sqrt{d+g}}{1 - \sqrt{d+g}} \right)
    -
     \frac{2 a (d - g) + c}{\sqrt{d - g}} \ln \left( \frac{1 + \sqrt{d-g}}{1 - \sqrt{d-g}} \right)
    \right],
$$
which can then be rewritten in terms of $\tan^{-1}$ or $\tanh^{-1}$ depending on the properties of the quantitiy $\sqrt{d \pm g}$.
