Integral by the residue theorem 
Check that the value of
  $$I(\alpha)=28\int_{-\infty}^\infty \frac{1-x^2}{2+3x^2+2\alpha^2x^4}\text{d}x$$
  is
  $$I(\alpha)=14\pi\left(1-\frac{1}{\alpha}\right)\sqrt{\frac{2}{4\alpha+3}}$$
  for some $\alpha>0$.

I've tried to apply the Residue Theorem to this integral with the contour $C_R\cup C$, where $C_R$ is the semicircle of radius $R$ centered at the origin 
(with initial point $(R,0)$ and final point $(-R,0)$) and $C$ is the segment $(-R,R)$ (with initial point $(-R,0)$ and final point $(R,0)$).
$\textbf{Lemma}$. Suppose $f(z)$ is defined in the upper half-plane. If there is an $a > 1$ and $M > 0$ such that $|f(z)| <\frac{M}{|z|^a}$ for large $|z|$. Then,
$$\lim_{R\to \infty} \int_{C_R} f(z)\text{d}z=0$$
The roots of the polynomial $p(x)=2+3x^2+2\alpha^2x^4$ are
$$x_1=-\frac{1}{2}\sqrt{\frac{-\sqrt{9-16\alpha^2}-3}{\alpha^2}}, \qquad x_2=\frac{1}{2}\sqrt{\frac{-\sqrt{9-16\alpha^2}-3}{\alpha^2}}$$
$$x_3=-\frac{1}{2}\sqrt{\frac{\sqrt{9-16\alpha^2}-3}{\alpha^2}}, \qquad x_4=\frac{1}{2}\sqrt{\frac{\sqrt{9-16\alpha^2}-3}{\alpha^2}}$$
By the previous lemma and the residue theorem, we have that 
$$I(\alpha)=\int_{-\infty}^\infty \frac{28(1-x^2)}{2+3x^2+2\alpha^2x^4}\text{d}x=\lim_{R\to \infty} \int_C \frac{28(1-x^2)}{2+3x^2+2\alpha^2x^4}\text{d}x=2\pi i \sum_{i=1}^n\text{Res}(f,x_i)$$
However, I don't know how to discuss what roots are in the upper half-plane due to the parameter $\alpha$. 
Can anyone continue from here or give some hints to follow? 

You are on the right track, and only needs a bit of extra input to get to the answer. Assume for a moment that $\alpha > 3/4$. Then the zeros of $2+3t+3\alpha^2 t^2 = 0$ are
$$ t = \frac{-3 \pm i\sqrt{16\alpha^2-9}}{4a^2} = \left( \frac{\sqrt{4\alpha-3} \pm i\sqrt{4\alpha+3}}{\sqrt{8}\alpha} \right)^2. $$
Now, among the square roots of these values, the only ones with positive imaginary parts are
$$ x_{(1)} = \frac{\sqrt{4\alpha-3} + i\sqrt{4\alpha+3}}{\sqrt{8}\alpha} \qquad \text{and} \qquad x_{(2)} = \frac{-\sqrt{4\alpha-3} + i\sqrt{4\alpha+3}}{\sqrt{8}\alpha}. $$
Then, after some painful algebra,
$$ I(\alpha) = 2\pi i \sum_{k=1}^{2} \operatorname{Res}(f, x_{(k)}) = 14\pi \left(1 - \frac{1}{\alpha}\right) \sqrt{\frac{2}{4\alpha+3}}. $$
At this point, this equality is established only for $\alpha > 3/4$. But since both sides are analytic near $(0, \infty)$, it extends to all of $(0, \alpha)$ by the principle of analytic continuation.
