Arguing away/bounding an integral in contour integration basically I have a contour integration problem where I am trying to determine the value of
 $$\int \frac {(z^2-1)^{1/2}} {(z^2+1)} dz$$ around a dumbbell contour. I am computing this integral with the ultimate aim of finding  $$\int_{-1}^{1} \frac {(1-x^2)^{1/2}} {(x^2+1)} dx$$

In my problem I need to "argue away" the components of my integral around two circles of small radius approaching zero, i.e. show these components tend to $0$ as the radius tends to $0$

We have on a circle $\gamma$ of radius $\epsilon$ centred at 1 

$$\vert \int \frac {(z^2-1)^{1/2}} {(z^2+1)} dz \vert \le 2\pi\epsilon .\max \vert \frac {(z^2-1)^{1/2}} {(z^2+1)} \vert$$

It is the $\epsilon\max \vert \frac {(z^2-1)^{1/2}} {(z^2+1)} \vert = \epsilon\frac {\max \vert (z^2-1)^{1/2} \vert}{\min \vert z^2+1 \vert}$ on $\gamma$ that I'm struggling to find. I need to show this is of order $\epsilon$ to a positive power and hence as $\epsilon \to 0$ so too does the integral. I was wondering if anyone could help?
You have to imagine that, attached to your dumbbell is a circle of radius $R$ centered at the origin.  Thus, letting all of the radii of the small detours about $z=\pm1$ go to zero, the contour integral is equal to

$$\oint_C dz \frac{(z^2-1)^{1/2}}{z^2+1} = i 2 \int_{-1}^1 dx \frac{(1-x^2)^{1/2}}{1+x^2} + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{(R^2 e^{i 2 \theta}-1)^{1/2}}{R^2 e^{i 2 \theta}+1}$$

As $R \to \infty$, the latter integral about the big circle approaches $i 2 \pi$.

By the residue theorem, the contour integral is equal to $i 2 \pi$ times the sum of the residues at the poles of the integrand at $z=\pm i$.  Thus,

$$i 2 \int_{-1}^1 dx \frac{(1-x^2)^{1/2}}{1+x^2} + i 2 \pi = i 2 \pi \left (\frac{\sqrt{2} e^{i \pi/2}}{2 e^{i \pi/2}} + \frac{\sqrt{2} e^{-i \pi/2}}{2 e^{-i \pi/2}} \right )$$

or,


  $$\int_{-1}^1 dx \frac{(1-x^2)^{1/2}}{1+x^2} = (\sqrt{2}-1) \pi$$


ADDENDUM

In my haste I did not see that the OP's concern is those small circles.  Consider $z=-1$.  Above the real axis, we parametrize by $z=e^{i \pi}+\epsilon \, e^{i \phi}$ and have the contribution to the contour integral being

$$i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{(2 e^{i \pi} \epsilon e^{i \phi})^{1/2}}{2} $$

which vanishes as $\epsilon \to 0$.  Same goes below the axis.  Similar analysis for $z=-1$, except we only need one integral from $\pi$ to $-\pi$.
You have to imagine that, attached to your dumbbell is a circle of radius $R$ centered at the origin.  Thus, letting all of the radii of the small detours about $z=\pm1$ go to zero, the contour integral is equal to

$$\oint_C dz \frac{(z^2-1)^{1/2}}{z^2+1} = i 2 \int_{-1}^1 dx \frac{(1-x^2)^{1/2}}{1+x^2} + i R \int_{-\pi}^{\pi} d\theta \, e^{i \theta} \frac{(R^2 e^{i 2 \theta}-1)^{1/2}}{R^2 e^{i 2 \theta}+1}$$

As $R \to \infty$, the latter integral about the big circle approaches $i 2 \pi$.

By the residue theorem, the contour integral is equal to $i 2 \pi$ times the sum of the residues at the poles of the integrand at $z=\pm i$.  Thus,

$$i 2 \int_{-1}^1 dx \frac{(1-x^2)^{1/2}}{1+x^2} + i 2 \pi = i 2 \pi \left (\frac{\sqrt{2} e^{i \pi/2}}{2 e^{i \pi/2}} + \frac{\sqrt{2} e^{-i \pi/2}}{2 e^{-i \pi/2}} \right )$$

or,


  $$\int_{-1}^1 dx \frac{(1-x^2)^{1/2}}{1+x^2} = (\sqrt{2}-1) \pi$$


ADDENDUM

In my haste I did not see that the OP's concern is those small circles.  Consider $z=-1$.  Above the real axis, we parametrize by $z=e^{i \pi}+\epsilon \, e^{i \phi}$ and have the contribution to the contour integral being

$$i \epsilon \int_{\pi}^0 d\phi \, e^{i \phi} \frac{(2 e^{i \pi} \epsilon e^{i \phi})^{1/2}}{2} $$

which vanishes as $\epsilon \to 0$.  Same goes below the axis.  Similar analysis for $z=-1$, except we only need one integral from $\pi$ to $-\pi$.