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

Here, we express the denominator in terms of sin x - cos x which is the integration of numerator.

Clearly,

$$\begin{gathered} (\sin \mathrm{x} - \cos \mathrm{x}{)}^2 = {\sin }^2 \mathrm{x} + {\cos }^2 \mathrm{x} - 2 \sin \mathrm{xcos} \mathrm{x} \\ = 1-\sin 2\mathrm{x} \\ \Rightarrow \sin 2 \mathrm{x} = 1 - (\sin \mathrm{x} - \cos \mathrm{x}{)}^2 \\ \therefore \mathrm{I} = {\int }_0^{\mathrm{\pi }/4}\frac{\sin \mathrm{x} + \cos \mathrm{x}}{16 + 9 \left\{1 - (\sin \mathrm{x} - \cos \mathrm{x}{)}^2\right\}} \mathrm{dx} \\ \Rightarrow \mathrm{I} = {\int }_0^{\mathrm{\pi }/4}\frac{\sin \mathrm{x} + \cos \mathrm{x}}{25-9 (\sin \mathrm{x} - \cos \mathrm{x}{)}^2} \mathrm{dx} \\ \mathrm{Let} \sin \mathrm{x} - \mathrm{cosx} = \mathrm{t}, \\ \mathrm{Then} , \mathrm{d} (\sin \mathrm{x} - \cos \mathrm{x}) = \mathrm{dt} \\ \Rightarrow (\cos \mathrm{x} + \sin \mathrm{x} ) \mathrm{dx} = \mathrm{dt} \\ \mathrm{also}, \mathrm{x} = 0 \\ \Rightarrow \mathrm{t} = \sin 0 - \cos 0 = - 1 \\ \mathrm{and} \mathrm{x} = \frac{\mathrm{\pi }}{4} \\ \mathrm{t} = \sin \frac{\mathrm{\pi }}{4} - \cos \frac{\mathrm{\pi }}{4} = 0 \\ \therefore \mathrm{I}= {\int }_{-1}^0 \frac{\mathrm{dt}}{25-9{\mathrm{t}}^2} \\ = \frac{1}{9}{\int }_{-1}^0 \frac{\mathrm{dt}}{{\displaystyle \frac{25}{9}-{\mathrm{t}}^2}} \\ =\frac{1}{9} {\int }_{-1}^0\frac{\mathrm{dt}}{{\left({\displaystyle \frac{5}{3}}\right)}^2-{\mathrm{t}}^2} \\ \Rightarrow \mathrm{I} = \frac{\mathrm{I}}{9} \mathrm{x} \frac{1}{2 (5/3)} {\left[\log \left|\frac{{\displaystyle \raisebox{1ex}{$5$}\!\left/ \!\raisebox{-1ex}{$3$}\right. + \mathrm{t}}}{5/3 - \mathrm{t}}\right|\right]}_{-1}^0 \\ \Rightarrow \mathrm{I}=\frac{1}{30}\left[\log - 1 \log \left(\frac{2/3}{8/3}\right) \right] \\ = \frac{1}{30} \left[\log 1 - \log \left(\frac{1}{4}\right)\right] \\ = \frac{1}{30} [ \log 1 + \log 4 ] \\ = \frac{1}{30} \log 4 = \frac{1}{15} \log 2 \end{gathered}$$