I have the following integral $$I=\int_a^\infty \frac{dx}{x^2-\frac{1}{x^2}},$$ where $a>0$. This integral contains a pole at $x=1$, and I rewrite the integral as $$\int_a^\infty \frac{x^2}{x^4-1}dx=\int_a^\infty \frac{x^2}{(x+1)(x^2+1)(x-1)}dx.$$
I'm not sure how deforming the contour works in this case with finite limits. My attempt was to break up the integral as $$I=\text{lim}_{\epsilon\to0}\left(\int_a^{1-\epsilon} \frac{x^2}{x^4-1}dx+\int_{1+\epsilon}^{\infty} \frac{x^2}{x^4-1}dx\right),$$ analytically extend the function to a complex function $f(x)\to f(z),\,z\in\mathbb{C}$, and then take a contour that goes as follows $$I_C=I+I_R+I_{\epsilon},$$ where $I_R$ is a big arc extending in the complex z plane connecting $a$ and $\infty$, $I_\epsilon$ is a small arc connecting $1-\epsilon$ and $1+\epsilon$, and $I$ is composed of the straight line segments from $a$ to $1-\epsilon$ and $1+\epsilon$ to $\infty$.
If the contour encircles the pole, we will have that $$I_C=2\pi i\,\text{Res}\bigg[\frac{z^2}{z^4-1}\bigg]=\frac{i\pi}{2}.$$ If I parametrise the $\epsilon$ arc by $z-1=re^{i\theta}$, the small arc integral is
$$I_\epsilon=\frac{i\pi}{4}.$$
Parametrising the big arc as $z=Re^{i\phi}$, the integral scales like $\frac{1}{R}$ and so the integral over the big arc vanishes. I'm not certain about this step since we have finite limits.
Therefore I'm left with $$\frac{i\pi}{2}=I+0+\frac{i\pi}{4}\rightarrow\,I=\frac{i\pi}{4}.$$
However this isn't the answer given in my source, there they say that for $a<1$ one gets $$I=\frac{1}{2}\left(\text{tanh}^{-1}a+\frac{\pi}{2}-\text{tan}^{-1}a-\frac{i\pi}{4}\right).$$
I don't understand how they reach this answer.
Update: If I calculate the principal value of the integral, I find $$\text{lim}_{\epsilon\to0}\left(\int_a^{1-\epsilon} \frac{x^2}{x^4-1}dx+\int_{1+\epsilon}^{\infty} \frac{x^2}{x^4-1}dx\right)=\frac{1}{4}\left(-2\text{tan}^{-1}a+2\text{tanh}^{-1}a+\pi\right),$$
which is almost the answer, it's just missing the $\frac{-i\pi}{4}$, so what am I missing in this calculation?