Let 2^{1-a} + 2^{1+a}, f(a), 3^a + 3^{-a} be | vedclass

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(B) Given that $2^{1-a} + 2^{1+a}$, $f(a)$, and $3^a + 3^{-a}$ are in $A$.$P$.,we have $2f(a) = (2^{1-a} + 2^{1+a}) + (3^a + 3^{-a})$.Using the $AM$-$

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$\frac{1}{4}\log_e\left(\frac{4}{3}\right)$

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(B) Given that $2^{1-a} + 2^{1+a}$, $f(a)$, and $3^a + 3^{-a}$ are in $A$.$P$.,we have $2f(a) = (2^{1-a} + 2^{1+a}) + (3^a + 3^{-a})$.Using the $AM$-$GM$ inequality, $2^{1-a} + 2^{1+a} = 2(2^{-a} + 2^a) \geq 2(2) = 4$ and $3^a + 3^{-a} \geq 2$. Both reach their minimum at $a=0$.Thus, the minimum value $\alpha$ of $f(a)$ is $\alpha = \frac{1}{2}(4 + 2) = 3$.The integral becomes $I = \int_{\log_e(2)}^{\log_e(3)} \frac{dx}{e^{2x} - e^{-2x}} = \int_{\log_e(2)}^{\log_e(3)} \frac{e^{2x} dx}{e^{4x} - 1}$.Let $t = e^{2x}$, then $dt = 2e^{2x} dx$, so $e^{2x} dx = \frac{dt}{2}$.When $x = \log_e(2)$, $t = e^{2\log_e(2)} = 4$. When $x = \log_e(3)$, $t = e^{2\log_e(3)} = 9$.$I = \frac{1}{2} \int_{4}^{9} \frac{dt}{t^2 - 1} = \frac{1}{2} \cdot \frac{1}{2} [\log_e|\frac{t-1}{t+1}|]_{4}^{9} = \frac{1}{4} [\log_e(\frac{8}{10}) - \log_e(\frac{3}{5})] = \frac{1}{4} \log_e(\frac{8/10}{3/5}) = \frac{1}{4} \log_e(\frac{4}{3})$.

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 $\int_{\log_e(\alpha-1)}^{\log_e(\alpha)} \frac{dx}{e^{2x} - e^{-2x}}$ is:

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