<s>[INST] Suppose $aSuppose $a<b<c<d$ and $p(x)=(x-a)(x-b)(x-c)(x-d)$. Show that 
$$\int_a^b \frac{dx}{\sqrt{|p(x)|}} = \int_c^d \frac{dx}{\sqrt{|p(x)|}}.$$
My attempt: I perform linear substitution $u=x-a+c$ to change the limits but I'm unable to change it to the required limits. I will end up with the upper limit $b-a+c$, which requires $b+c=a+d$, which is not always true (I think). Can anyone guide me?
 [/INST] If $a,b,c,d$ has a center of symmetry $\frac{a+d}{2}= \frac{b+c}{2}$ it's clear. Otherwise, we could consider a "generalized symmetry", an involution $x\leftrightarrow x'$ given by a symmetric equation 
$$u\, x x' + v\, (x+x') + w =0$$
Let's consider this particular case $a,b,c,d= 1,2,4,7$.  We want $1\leftrightarrow 7 $ and $2\leftrightarrow 4$ which gets us $(u:v:w) = (2:1:22)$.
The equality $u \,x x' + v \,(x+x') + w =0$ solved for $x'$ gets us 
$$x' = \iota(x)= \frac{- v\, x  - w}{u\, x + v}$$
which in our case gives $\iota(x)= \frac{-x + 22}{2 x +1}$. We now check that all fits perfectly: the form $\frac{(d x)^2 }{p(x)}$ is invariant under the substitution $x \mapsto \iota(x)$. Indeed, we have
\begin{eqnarray}
\frac{d \iota(x)}{dx}&=& - \frac{45}{(1+2 x)^2}\\
 (\iota(x)-1)(\iota(x)-2)(\iota(x)-4)(\iota(x)-7) &=& \frac{2025}{(2 x+1)^4} \cdot (x-1)(x-2)(x-4)(x-7)
\end{eqnarray}
both following from the  identity $\frac{\iota(x)-\iota(y)}{x-y} = -\frac{45}{(2x+1)(2y+1)}$
Hence $\frac{|dx|}{\sqrt{|p(x)|}}$ is invariant under $\iota$ and so
$$\int_1^2 \frac{dx}{\sqrt{|p(x)|}} = \int_1^2 \frac{|d\iota(x)|}{\sqrt{|p(\iota(x))|}}= \int_4^7 \frac{dx}{\sqrt{|p(x)|}} $$
$\bf{Added:}$
Some numerics: $\int_1^2\frac{dx}{\sqrt{-(x-1)(x-2)(x-4)(x-7)}}=0.8570..=\int_4^7\frac{dx}{\sqrt{-(x-1)(x-2)(x-4)(x-7)}} $.
What about $\int_2^4\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}}$ ? It's $1.1656..$. But note that $\int_{-\infty}^1\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}}=0.72316..$ and 
$\int_7^{\infty}\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}}=0.44245..$ so 
$$\int_2^4\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}} = \int_{-\infty}^1\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}}+ \int_7^{\infty}\frac{dx}{\sqrt{(x-1)(x-2)(x-4)(x-7)}}$$
the last sum should be thought of as an integral from $1$ to $7$ going around through $\infty$. The proof uses a similar involution. 
</s>