Let ∫ x ^ 3 + x ^ 2 + x 12 x ^ 3 + 15 x ^ 2 + 20x dx = f(x) where...

In the given integral, multiplying numerator and denominator by $x$, we get,

$f\left(x\right) = ∫ \dfrac{{x}^{4} + {x}^{3} + {x}^{2}}{\sqrt{12{x}^{5} + 15{x}^{4} + 20{x}^{3}}}dx$
Now substitute $12{x}^{5} + 15{x}^{4} + 20{x}^{3} = {t}^{2}$
$⇒ 60\left({x}^{4} + {x}^{3} + {x}^{2}\right)dx = 2tdt$
$⇒ f\left(x\right) = \dfrac{1}{30} ∫ \dfrac{tdt}{t} = \dfrac{1}{30}\sqrt{12{x}^{5} + 15{x}^{4} + 20{x}^{3}} + C$
As $f\left(1\right) = \dfrac{\sqrt{47}}{30} ⇒ C = 0$
So, $f\left(2\right) = \dfrac{\sqrt{784}}{30} ⇒ {\left(f\left(2\right)\right)}^{2} = \dfrac{784}{900} = \dfrac{196}{225}$

Hence, $p = 196$