The value $\underset{\sqrt{\ln 2}}{\overset{\sqrt{\ln 3}}{\int }}\frac{x\sin x^2}{\sin x^2+\sin (\ln 6-x^2)}dx$ is

$\underset{\sqrt{\ln 2}}{\overset{\sqrt{\ln 3}}{\int }}\frac{x\sin x^2}{\sin x^2+\sin (\ln 6-x^2)}dx$

$put{x}^2=t\Rightarrow LL:x=\sqrt{\ln 2}\Rightarrow t=\ln 2$

$UL:x=\sqrt{\ln 3}\Rightarrow t=\ln 3$

$=\frac{1}{2}\underset{ln2}{\overset{ln3}{\int }}\frac{\sin t}{\sin t+\sin (\ln 6-t)}dt$

$=\frac{1}{2}.\frac{1}{2}(\ln 3-\ln 2)=\frac{1}{4}\ln \frac{3}{2}$

$\left(\because \underset{a}{\overset{b}{\int }}\frac{f\left(x\right)}{f\left(x\right)+f(a+b-x)}dx=\frac{(b-a)}{2}\right)$