How to integrate $\int_{0}^{1} \frac{1-x}{1+x} \frac{dx}{\sqrt{x^4 + ax^2 + 1}}$?

Enforcing the substitution $x^{-1}-x=u$ in your last integral we get:

$$ \int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+u^2}}\right)\frac{du}{u\sqrt{u^2+a+2}} $$ and by setting $u=\sqrt{a+2}\sinh\theta$ we get: $$ \frac{1}{\sqrt{a+2}}\int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+(a+2)\sinh^2\theta}}\right)\frac{d\theta}{\sinh\theta}$$ We may get rid of the last term through the "hyperbolic Weierstrass substitution" $$ \theta = 2\,\text{arctanh}(e^{-v}) = \log\left(\frac{e^v+1}{e^v-1}\right)$$ that wizardly gives $$ \frac{1}{\sqrt{a+2}}\int_{0}^{+\infty}\left(-1+\frac{2}{\sqrt{4+\frac{a+2}{\sinh^2 v}}}\right)\,dv$$ i.e., finally, a manageable integral through differentiation under the integral sign.
This proves OP's initial identity. Beers on me.

$$\text{let } \ \frac{1-x}{1+x}=t \Rightarrow x=\frac{1-t}{1+t}\Rightarrow dx=-\frac{2}{(1+t)^2}dt\ \text{ then:}$$ $$\int_0^1 \frac{1-x}{1+x}\frac{dx}{\sqrt{x^4+ax^2+1}}=\int_0^1 \frac{2t}{\sqrt{(a+2)t^4-2(a-6)t^2+(a+2)}}dt$$ $$\overset{t^2=x}=\frac{1}{\sqrt{a+2}}\int_0^1 \frac{dx}{\sqrt{x^2-2\left(\frac{a-6}{a+2}\right)x+1}}=\boxed{\frac{1}{\sqrt{a+2}}\ln \left(1+\frac{\sqrt{a+2}}{2}\right),\quad a>-2}$$