If only you could flip it...

Rewrite the integral like this. $\int_0^3\dfrac{2x^3}{x^2+9}\text{ }dx=\int_0^3\dfrac{2x^3+18x-18x}{x^2+9}\text{ }dx$ This splits into two integrals that are easy to solve. $\int_0^3\dfrac{2x^3+18x-18x}{x^2+9}\text{ }dx=\int_0^3\dfrac{2x^3+18x}{x^2+9}\text{ }dx-\int_0^3\dfrac{18x}{x^2+9}\text{ }dx$ Solve for the indefinite integrals and then plug in $0$ and $3.$

The first integral simplifies to this. $\int2x\text{ }dx=x^2+C$ The second integral can be solved with a $u\text{-substitution}$ for $x^2+9.$ $\int\dfrac{18x}{x^2+9}\text{ }dx=9\ln(x^2+9)+C$ So overall, the integral is equal to this. $\int\dfrac{2x^3+18x}{x^2+9}\text{ }dx-\int\dfrac{18x}{x^2+9}\text{ }dx=x^2-9\ln(x^2+9)+C$ Now plug in $0$ and $3.$ $\int_0^3\dfrac{2x^3+18x}{x^2+9}\text{ }dx-\int_0^3\dfrac{18x}{x^2+9}\text{ }dx=\left.x^2-9\ln(x^2+9)\right]^3_0$ This is equal to $9-9\ln18-0+\ln9=9-9\ln2.$ Therefore, $A=9,$ $B=9,$ and $C=2.$ $A+B+C=\boxed{20}$

put z= (x) ^ 2 ......... rest is simple integration!

Substituting $x^{2} + 9 = t$ $x^{2} = t-9$ Then $2xdx = dt$ $2x^{3}dx = x^{2}dt$ $2x^{3}dx = (t-9)dt$ $\int _{ 0 }^{ 3 }{ \frac { 2{ x }^{ 3 } }{ { x }^{ 2 }+9 } \quad =\quad \int _{ 9 }^{ 18 }{ \frac { t-9 }{ t } } } dt\quad =\quad { \left[ t-9\ln { t } \right] }_{ 9 }^{ 18 }\\$ Solving we get integral = 9 - 9ln2; A=9; B=9; C=2 Hence A +B+ C = 20