To solve the integral \(\int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin x^2}{\sin x^2 + \sin(\ln 6 - x^2)} \, dx\), we start by observing a possible symmetry in the function within the integral.
Let's consider the substitution \( u = \ln 6 - x^2 \). Then \( du = -2x \, dx \) or \( x \, dx = -\frac{1}{2} \, du \).
Under this substitution, when \( x = \sqrt{\ln 2} \), \( u = \ln 6 - \ln 2 = \ln \frac{6}{2} = \ln 3 \).
Similarly, when \( x = \sqrt{\ln 3} \), \( u = \ln 6 - \ln 3 = \ln \frac{6}{3} = \ln 2 \).
The limits of integration get reversed due to negative sign from substitution, changing the integral:
\(\int_{\ln 3}^{\ln 2} \frac{-\frac{1}{2} \sin (\ln 6 - u)}{\sin (\ln 6 - u) + \sin u} \, du\)
Reversing the limits, we change the sign of the integral:
\(\int_{\ln 2}^{\ln 3} \frac{\frac{1}{2} \sin (\ln 6 - u)}{\sin (\ln 6 - u) + \sin u} \, du\)
By symmetry, adding the original integral and this transformed integral gives:
\(\int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin x^2}{\sin x^2 + \sin(\ln 6 - x^2)} \, dx + \int_{\ln 2}^{\ln 3} \frac{\frac{1}{2} \sin u}{\sin u + \sin(\ln 6 - u)} \, du = \frac{1}{2}(\ln 3 - \ln 2)\)
This implies:
\(2\int_{\sqrt{\ln 2}}^{\sqrt{\ln 3}} \frac{x \sin x^2}{\sin x^2 + \sin(\ln 6 - x^2)} \, dx = \frac{1}{2} \ln \frac{3}{2}\)
Thus, the value of the original integral is:
\(=\frac{1}{4} \ln \frac{3}{2}\)
Therefore, the correct answer is \(\frac{1}{4} \ln \frac{3}{2}\).