Differentiation seems easy ?

(This is only a $\textbf{HINT}$ ):

$x^4-x^2+1$ = $(x^2+1-\sqrt{3}x)(x^2+1+\sqrt{3}x)$

To solve this question, we first need to differentiate $y$ with respect to $x$ .

$y=\cfrac{x^{4}-x^{2}+1}{x^2+\sqrt{3}x+1}$

$\boxed{\cfrac{d}{dx} \cfrac{u}{v}=\cfrac{v\cfrac{du}{dx}-u\cfrac{dv}{dx}}{v^{2}}}$

$\cfrac{dy}{dx}=\cfrac{d}{dx}\cfrac{x^{4}-x^{2}+1}{x^2+\sqrt{3}x+1}$

$=\cfrac{(x^2+\sqrt{3}x+1)\cfrac{d}{dx}(x^{4}-x^{2}+1)-(x^{4}-x^{2}+1)\cfrac{d}{dx}(x^2+\sqrt{3}x+1)}{(x^2+\sqrt{3}x+1)^{2}}$

$=\cfrac{(x^2+\sqrt{3}x+1)(4x^{3}-2x)-(x^{4}-x^{2}+1)(2x+\sqrt{3})}{(x^2+\sqrt{3}x+1)^{2}}$

$=2x-\sqrt{3}$

Now, given: $\cfrac{dy}{dx}=ax+b$

So, $2x-\sqrt{3}=ax+b$

On comparing, we get: $a=2$ , $b=-\sqrt{3}$

Now, $a+b=c-\sqrt{d}$

So, $2-\sqrt{3}=c-\sqrt{d}$

On comparing, we get: $c=2$ , $d=3$

So, $c+d=5$

Thus, the answer is: $\boxed{c+d=5}$

$y = \frac{ax^2}{2}+bx+c = \frac{x^4 - x^2 + 1}{x^2+\sqrt{3}x+1}$ .

$\lim_{x\to\infty} y = \frac{ax^2}{2} = x^2$ , so $a = 2$ .

$y(0) = c = 1$ .

$y(-\sqrt{3}) = 3 -\sqrt{3}b + 1 = \frac{9 - 3 + 1}{1} = 7$ , so $b = -\sqrt{3}$ .

$c+d = 5$ .

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I did it via evaluating $y$ , but thought I'd post an alternative solution.

note that x^4-x^2+1=(x^2+1+sqart(3).x)(x^2+1-sqart(3).x) so
y=(x^2-sqart(3).x+1) and y`=2x-sqrt(3). hence c=2 and d=3 and c+d=5