Let $I=\int_{0}^{\frac{\pi}{4}} \frac{\sin x+\cos x}{9+16 \sin 2 x} d x=K \log 3$
Put $\sin x-\cos x=t \Rightarrow(\cos x+\sin x) d x=d t$ ...(1)
Now squaring equation (1), we get
$\begin{aligned} \therefore I &=\int_{-1}^{0} \frac{d t}{9+16\left(1-t^{2}\right)} \\ &=\int_{-1}^{0} \frac{d t}{25-16 t^{2}}=\frac{1}{16} \int_{-1}^{0} \frac{d t}{\left(\frac{5}{4}\right)^{2}-t^{2}} \end{aligned}$
$\begin{array}{l}
=\frac{1}{16} \times \frac{1}{2\left(\frac{5}{4}\right)}\left[\log \mid \frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right]_{-1}^{0}=\frac{1}{40}\left[\log \left(\frac{5+4 t}{5-4 t}\right)\right]_{-1}^{0}=\frac{1}{40}\left[\log (1)-\log \left(\frac{1}{9}\right)\right] \\
=\frac{1}{40}(\log 9)=\frac{1}{40} \log 3^{2}=\frac{2}{40} \log 3=\frac{1}{20} \log 3
\end{array}$
As per given data, $K=\frac{1}{20}$