Lol i think this method is a complete unintended tedious one
use the substitution x=sqrt(3)sin(theta). the new limits are now 0 to sin^-1(1/sqrt(3)). and dx=sqrt(3)cos(theta). Sub it in, the integrand becomes 9sin^2(theta)cos^2(theta). Which can be written as 9/4(sin2theta)^2. Using the double angle cos formula, it will turn out to be 9/8(1-cos(4theta)). Now integrating it, it becomes 9/8(theta-(sin4theta)/4)) from 0 to sin^-1(1/sqrt3). Expanding out the sin4theta (either by continuous use of double angle formula or DMT) and subbing in the limits will give a final answer of 1/8(9arcsin(1/sqrt3)-sqrt2).
soz for no latex might add it when im free