Consider$\int \frac{3x^4+2x^2+1}{\sqrt{x^4+x^2+1}}dx=f\left(x\right)$. If $f\left(1\right)=\sqrt{3}$, then ${\left(f\left(2\right)\right)}^2$ is equal to

Multiplying the numerator and denominator by $x$ in the given integral, we get,

$f\left(x\right)=\int \frac{3x^5+2x^3+x}{\sqrt{x^6+x^4+x^2}}dx$ 

Now substitute $x^6+x^4+x^2=t^2$
$\Rightarrow 2\left(3x^5+2x^3+x\right)dx=2tdt$
So, $f\left(x\right)=\int \frac{tdt}{t}=\sqrt{x^6+x^4+x^2}+C$
As $f\left(1\right)=\sqrt{3}\Rightarrow C=0$

Hence, $f\left(2\right)=\sqrt{64+16+4}\Rightarrow {\left(f\left(2\right)\right)}^2=84$