Consider ∫ 3 x ^ 4 + 2 x ^ 2 + 1 x ^ 4 + x ^ 2 + 1 dx = f(x) . If...

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

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

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

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