1. Let $I=\int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x\sin x^2}{\sin x^2+\sin(\ln 6-x^2)}\,dx.$

We simplify the integral by substituting $t=x^2 \implies dt=2x\,dx \implies x\,dx=\frac{dt}{2}.$

Also, the limits change as: $x=\sqrt{\ln 2} \Rightarrow t=\ln 2,$ $x=\sqrt{\ln 3} \Rightarrow t=\ln 3.$

So, $I=\frac12\int_{\ln 2}^{\ln 3} \frac{\sin t}{\sin t+\sin(\ln 6-t)}\,dt.$

2. Now observe a symmetry. Let $f(t)=\frac{\sin t}{\sin t+\sin(\ln 6-t)}.$ Then, $f(\ln 6-t)=\frac{\sin(\ln 6-t)}{\sin(\ln 6-t)+\sin t}.$ Hence, $f(t)+f(\ln 6-t)=1.$

3. Also note that the interval $[\ln 2,\ln 3]$ is symmetric with respect to the transformation $t\mapsto \ln 6-t$, because: $\ln 6-\ln 2=\ln 3,$ $\ln 6-\ln 3=\ln 2.$

Therefore, $\int_{\ln 2}^{\ln 3} f(t)\,dt=\int_{\ln 2}^{\ln 3} f(\ln 6-t)\,dt.$ Adding the two, 2\int_{\ln 2}^{\ln 3} f(t)\,dt=int_{\ln 2}^{\ln 3} [f(t)+f(\ln 6-t)]\,dt=int_{\ln 2}^{\ln 3} 1\,dt. So, $2\int_{\ln 2}^{\ln 3} f(t)\,dt=\ln 3-\ln 2=\ln\frac32.$ Hence, $\int_{\ln 2}^{\ln 3} f(t)\,dt=\frac12\ln\frac32.$

4. Therefore, $I=\frac12\cdot \frac12\ln\frac32=\frac14\ln\frac32.$

5. Compare with options:

• A: $\frac14\ln\frac32$ ✅
• B: $\frac12\ln\frac32$
• C: $\ln\frac32$
• D: $\frac16\ln\frac32$

So the correct option is A.