If $\alpha, \beta(\beta>\alpha)$ are the roots of $f(x) \equiv a x^2+b x+c=0$ and $f(x)$ is an even function, then

$\int\limits_\alpha^\beta \frac{e^{f\left(\frac{f(x)}{x-\alpha}\right)}}{e^{f\left(\frac{f(x)}{x-\alpha}\right)}+e^{f\left(\frac{f(x)}{x-\beta}\right)}} d x$ is equal to

$\frac{\sqrt{b^2-4 a c}}{|2 a|}$

It is given that $f(x)$ is an even function. Therefore,

$f(-x)=f(x)$ for all $x \in R$

$\Rightarrow a x^2-b x+c=a x^2+b x+c$ for all $x \in R$

$\Rightarrow b=0 \Rightarrow \frac{-b}{a}=0 \Rightarrow \alpha+\beta=0 \Rightarrow \beta=-\alpha$

Also, $\frac{f(-x)}{-x-\alpha}=-\frac{f(x)}{x-\beta}$ and $\frac{f(-x)}{-x-\beta}=\frac{f(x)}{-x+\alpha}=-\frac{f(x)}{x-\alpha}$

∴  $f\left(\frac{f(-x)}{-x-\alpha}\right)=f\left(-\frac{f(x)}{x-\beta}\right)=f\left(\frac{f(x)}{x-\beta}\right)$

and, $f\left(\frac{f(-x)}{-x-\beta}\right)=f\left(-\frac{f(x)}{x-\alpha}\right)=f\left(\frac{f(x)}{x-\alpha}\right)$

Let $I=\int\limits_\alpha^\beta \frac{e^{f\left(\frac{f(x)}{x-\alpha}\right)}}{e^{f\left(\frac{f(x)}{x-\alpha}\right)}+e^{f\left(\frac{f(x)}{x-\beta}\right)}} d x$         ......(i)

Using $\int\limits_a^b f(x) d x=\int\limits_a^b f(a+b-x) d x$ and $\alpha+\beta=0$, we obtain

$I =\int\limits_\alpha^\beta \frac{e^{f\left(\frac{f(-x)}{-x-\alpha}\right)}}{e^{f\left(\frac{f(-x)}{-x-\alpha}\right)}+e^{f\left(\frac{f(-x)}{-x-\beta}\right)}} d x$

$\Rightarrow I =\int\limits_\alpha^\beta \frac{e^{f\left(\frac{f(x)}{x-\beta}\right)}}{e^{f\left(\frac{f(x)}{x-\beta}\right)}+e^{f\left(\frac{f(x)}{x-\alpha}\right)}} d x$        ......(ii)

Adding (i) and (ii), we obtain

$2 I=\int\limits_\alpha^\beta d x=\beta-\alpha$

$\Rightarrow I=\frac{1}{2} \sqrt{(\beta+\alpha)^2-4 \alpha \beta}=\frac{\sqrt{b^2-4 a c}}{2|a|}$