<s>[INST] How to integrate $\int_{0}^{1} \frac{1-x}{1+x} \frac{dx}{\sqrt{x^4 + ax^2 + 1}}$? The question is how to show the identity
$$ \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} = \frac{1}{\sqrt{a+2}} \log\left( 1 + \frac{\sqrt{a+2}}{2} \right), \tag{$a>-2$} $$
I checked this numerically for several cases, but even Mathematica 11 could not manage this symbolically for general $a$, except for some special cases like $a = 0, 1, 2$.

Addendum. Here are some backgrounds and my ideas:


*

*This integral came from my personal attempt to find the pattern for the integral
$$ J(a, b) := \int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{1 + ax^2 + bx^4}}. $$
This drew my attention as we have the following identity
$$ \int_{0}^{\infty} \frac{x}{x+1} \cdot \frac{dx}{\sqrt{4x^4 + 8x^3 + 12x^2 + 8x + 1}} = J(6,-3), $$
where the LHS is the integral from this question. So establishing the claim in this question amounts to showing that $J(6,-3) = \frac{1}{2}\log 3 - \frac{1}{3}\log 2$, though I am skeptical that $J(a, b)$ has a nice closed form for every pair of parameters $(a, b)$.

*A possible idea is to write
\begin{align*}
&\int_{0}^{1} \frac{1-x}{1+x} \cdot \frac{dx}{\sqrt{x^4 + ax^2 + 1}} \\
&\hspace{5em}= \int_{0}^{1} \frac{(x^{-2} + 1) - 2x^{-1}}{x^{-1} - x} \cdot \frac{dx}{\sqrt{(x^{-1} - x)^2 + a + 2}}
\end{align*}
This follows from a simple algebraic manipulation. This suggests that we might be able to apply Glasser's master theorem, though in a less trivial way.
I do not believe that this is particularly hard, but I literally have not enough time to think about this now. So I guess it is a good time to seek help.
 [/INST] $$\int_0^1 \frac{1-x}{1+x}\frac{dx}{\sqrt{x^4+ax^2+1}}\overset{\large \frac{1-x}{1+x}=t}=\int_0^1 \frac{2t}{\sqrt{(a+2)t^4-2(a-6)t^2+(a+2)}}dt$$
$$\overset{t^2=x}=\frac{1}{\sqrt{a+2}}\int_0^1 \frac{dx}{\sqrt{x^2-2\left(\frac{a-6}{a+2}\right)x+1}}=\frac{1}{\sqrt{a+2}}\ln \left(1+\frac{\sqrt{a+2}}{2}\right)$$
</s>