Evaluate the definite integral $\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$.

(D) Let $I = \int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$.
Let $t = \sin x - \cos x$. Then $dt = (\cos x + \sin x) dx$.
When $x = 0$,$t = \sin 0 - \cos 0 = -1$. When $x = \frac{\pi}{4}$,$t = \sin \frac{\pi}{4} - \cos \frac{\pi}{4} = 0$.
Now,$t^2 = (\sin x - \cos x)^2 = \sin^2 x + \cos^2 x - 2 \sin x \cos x = 1 - \sin 2x$.
Therefore,$\sin 2x = 1 - t^2$.
Substituting these into the integral:
$I = \int_{-1}^{0} \frac{dt}{9 + 16(1 - t^2)} = \int_{-1}^{0} \frac{dt}{9 + 16 - 16t^2} = \int_{-1}^{0} \frac{dt}{25 - 16t^2}$.
Using the formula $\int \frac{dx}{a^2 - x^2} = \frac{1}{2a} \ln \left| \frac{a+x}{a-x} \right| + C$:
$I = \frac{1}{4} \int_{-1}^{0} \frac{dt}{5^2 - (4t)^2} = \frac{1}{4} \left[ \frac{1}{2(5)} \ln \left| \frac{5 + 4t}{5 - 4t} \right| \right]_{-1}^{0}$.
$I = \frac{1}{40} \left[ \ln \left| \frac{5+0}{5-0} \right| - \ln \left| \frac{5-4}{5+4} \right| \right] = \frac{1}{40} [ \ln(1) - \ln(1/9) ]$.
Since $\ln(1) = 0$ and $-\ln(1/9) = \ln(9)$:
$I = \frac{1}{40} \ln(9) = \frac{1}{40} \ln(3^2) = \frac{2}{40} \ln(3) = \frac{1}{20} \ln(3)$.