Let $$(2^{1-a} + 2^{1+a})$$, $$f(a)$$, $$(3^a + 3^{-a})$$ be in A.P. and $$\alpha$$ be the minimum value of  $$f(a)$$, Then the value of the integral $$\displaystyle\int_{\log_e(\alpha - 1)}^{\log_e(\alpha)} \frac{dx}{e^{2x} - e^{-2x}}$$ is equal to :

$$2f(a) = (2^{1-a} + 2^{1+a}) + (3^a + 3^{-a})$$

$$2^{1-a} + 2^{1+a} = \frac{2}{2^a} + 2 \cdot 2^a = 2\left(2^a + \frac{1}{2^a}\right)$$

$$2f(a) = 2\left(2^a + \frac{1}{2^a}\right) + \left(3^a + \frac{1}{3^a}\right)$$

$$f(a) = \left(2^a + \frac{1}{2^a}\right) + \frac{1}{2}\left(3^a + \frac{1}{3^a}\right)$$

$$\left(2^a + \frac{1}{2^a}\right) \ge 2$$

$$\left(3^a + \frac{1}{3^a}\right) \ge 2$$

$$\alpha = f(a)_{\min} = 2 + \frac{1}{2}(2) = 2 + 1 = 3$$

Thus, $$\alpha = 3$$ (at $$a=0$$)

$$I = \int_{\log_e 2}^{\log_e 3} \frac{dx}{e^{2x} - e^{-2x}}$$

$$I = \int_{\log_e 2}^{\log_e 3} \frac{e^{2x} \, dx}{(e^{2x})^2 - 1}$$

Let $$u = e^{2x} \implies du = 2e^{2x} \, dx \implies e^{2x} \, dx = \frac{du}{2}$$

$$I = \int_{4}^{9} \frac{\frac{du}{2}}{u^2 - 1} = \frac{1}{2} \int_{4}^{9} \frac{du}{u^2 - 1}$$

$$I = \frac{1}{2} \left[ \frac{1}{2} \log_e \left( \frac{u - 1}{u + 1} \right) \right]_{4}^{9}$$

$$I = \frac{1}{4} \left[ \log_e \left( \frac{9 - 1}{9 + 1} \right) - \log_e \left( \frac{4 - 1}{4 + 1} \right) \right]$$

$$I = \frac{1}{4} \left[ \log_e \left( \frac{8}{10} \right) - \log_e \left( \frac{3}{5} \right) \right] = \frac{1}{4} \left[ \log_e \left( \frac{4}{5} \right) - \log_e \left( \frac{3}{5} \right) \right]$$

$$I = \frac{1}{4} \log_e \left( \frac{4/5}{3/5} \right) = \frac{1}{4} \log_e \left( \frac{4}{3} \right)$$