$\int \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$ is equal to :

$\frac{1}{40} \ln \left|\frac{5+4(\sin x-\cos x)}{5-4(\sin x-\cos x)}\right|+c$

Let $I=\int \frac{\sin x+\cos x}{9+16 \sin 2 x} d x$

Let $t=\sin x-\cos x \Rightarrow t^2=1-\sin 2 x$

∴  $\sin 2 x=\left(1-t^2\right)$

∴  $I=\int \frac{d t}{9+16\left(1-t^2\right)}=\int \frac{d t}{25-16 t^2}$

$=\frac{1}{16} \int \frac{d t}{\left(\frac{5}{4}\right)^2-t^2}=\frac{1}{16} \times \frac{1}{2\left(\frac{5}{4}\right)} \ln \left|\frac{\frac{5}{4}+t}{\frac{5}{4}-t}\right|+c$

$=\frac{1}{40} \ln \left|\frac{5+4 t}{5-4 t}\right|+c=\frac{1}{40} \ln \left|\frac{5+4(\sin x-\cos x)}{5-4(\sin x-\cos x)}\right|+c$

Hence (1) is the correct answer.