The integral $$80\int_{0}^{\frac{\pi}{4}}\left(\frac{\sin \theta + \cos \theta}{9+16\sin 2\theta}\right)d\theta$$ is equaol to :

We need to evaluate $$80\int_{0}^{\pi/4}\frac{\sin\theta + \cos\theta}{9 + 16\sin 2\theta}\,d\theta$$. Noting that $$\sin 2\theta = 2\sin\theta\cos\theta$$ and $$(\sin\theta + \cos\theta)^2 = 1 + \sin 2\theta$$, we have $$\sin 2\theta = (\sin\theta + \cos\theta)^2 - 1$$.

To simplify the integrand, let $$t = \sin\theta - \cos\theta$$ so that $$dt = (\cos\theta + \sin\theta)\,d\theta$$. Also, since $$t^2 = 1 - \sin 2\theta$$, it follows that $$\sin 2\theta = 1 - t^2$$. When $$\theta = 0$$, $$t = -1$$, and when $$\theta = \pi/4$$, $$t = 0$$.

Under this substitution the integral becomes $$ 80\int_{-1}^{0}\frac{dt}{9 + 16(1 - t^2)} = 80\int_{-1}^{0}\frac{dt}{25 - 16t^2}. $$

Using partial fractions, $$ \frac{1}{25 - 16t^2} = \frac{1}{(5 - 4t)(5 + 4t)} = \frac{1}{10}\biggl(\frac{1}{5-4t} + \frac{1}{5+4t}\biggr). $$ Thus $$ 80 \times \frac{1}{10}\int_{-1}^{0}\Bigl(\frac{1}{5-4t} + \frac{1}{5+4t}\Bigr)dt = 8\Bigl[-\frac{1}{4}\ln|5-4t| + \frac{1}{4}\ln|5+4t|\Bigr]_{-1}^{0} = 2\Bigl[\ln\Bigl(\frac{5+4t}{5-4t}\Bigr)\Bigr]_{-1}^{0}. $$

Evaluating the logarithm gives $$ 2\Bigl[\ln\Bigl(\frac{5}{5}\Bigr) - \ln\Bigl(\frac{1}{9}\Bigr)\Bigr] = 2\ln 9 = 4\ln 3. $$ Hence the correct answer is Option 2: $$4\log_e 3$$.