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

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

$\frac{1}{9} \log |\frac{5 - 4 (\sin x - \cos x)}{5 + 4 (\sin x - \cos x)}| + c$

$\frac{1}{18} \log |\tan \frac{x}{2} + \sec \frac{x}{2}| + c$

$\frac{1}{8} \log |\frac{5 + 4 (\sin x - \cos x)}{5 - 4 (\sin x - \cos x)}| + c$

answer is C.

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Detailed Solution

$\begin{array}{l} \int \frac{\sin x + \cos x}{9 + 16 \sin 2 x} d x \\ = \int \frac{\sin x + \cos x}{25 - 16 (\sin x - \cos x)^{2}} d x \\ Where \sin 2 x = 1 - (\sin x - \cos x)^{2} \\ Put \sin x - \cos x = t \\ (\cos x + \sin x) d x = d t \\ = \int \frac{d t}{25 - 16 t^{2}} \\ = \frac{1}{16} \int \frac{d t}{{(\frac{5}{4})}^{2} - t^{2}} \end{array}$

$\begin{array}{l} = \frac{1}{16} \cdot \frac{1}{2 \cdot (\frac{5}{4})} \log |\frac{\frac{5}{4} + t}{\frac{5}{4} - t}| + c \\ = \frac{1}{40} \log |\frac{5 + 4 t}{5 - 4 t}| + c \\ = \frac{1}{40} \log |\frac{5 + 4 (\sin x - cox)}{5 - 4 (\sin x - \cos x)}| + c \end{array}$