Let $\int \frac{x^3+x^2+x}{\sqrt{12x^3+15x^2+20x}}dx=f\left(x\right)$ where $f\left(1\right)=\frac{\sqrt{47}}{30}.$ If ${\left(f\left(2\right)\right)}^2$ is equal to $\frac{p}{225},$ then the value of $p$ is equal to

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

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

Hence, $p=196$