If the value of the definite integral $I=\underset{1}{\overset{\infty }{\int }}\frac{\left(2x^3-1\right)dx}{x^6+2x^3+9x^2+1}$ can be expressed in the form $\frac{A}{B}{\cot }^{-1}\frac{C}{D}$ where $\frac{A}{B}$ and $\frac{C}{D}$ are rationals in their lowest form, then the value of $\left(A+B^2+C^3+D^4\right)$.

$I=\underset{1}{\overset{\infty }{\int }}\frac{\left(2x-x^{-2}\right)dx}{x^4+2x+9+x^{-2}}$(Dividing $Nr$ and $Dr$ by $x^2$)

Or   $I=\underset{1}{\overset{\infty }{\int }}\frac{\left(2x-x^{-2}\right)dx}{{\left(x^2+\frac{1}{x}\right)}^2+9}$

Put   $x^2+\frac{1}{x}=t$

$\Rightarrow \left(2x-\frac{1}{x^2}\right)dx=dt$

as   $x\to 1,t\to 2$

as    $x\to \infty ,t\to \infty$

$\therefore I=\underset{2}{\overset{\infty }{\int }}{\left.\frac{dt}{t^2+9}dx=\frac{1}{3}{\tan }^{-1}\frac{t}{3}\right]}_2^{\infty }$

$=\frac{1}{3}\left[\frac{\pi }{2}-{\tan }^{-1}\frac{2}{3}\right]=\frac{1}{3}{\cot }^{-1}\frac{2}{3}$

Hence, $A=1;B=3;C=2;D=3$

$\therefore \left(A+B^2+C^3+D^4\right)=1+9+8+81=99$