Evaluate: $\int\limits_{\log \sqrt{2}}^{\log \sqrt{3}} \frac{1}{(e^x + e^{-x})(e^x - e^{-x})} dx$

$\frac{1}{4} \log \frac{3}{2}$

The correct answer is Option (3) → $\frac{1}{4} \log \frac{3}{2}$

$I = \int\limits_{\log \sqrt{2}}^{\log \sqrt{3}} \frac{1}{(e^{2x} - e^{-2x})} dx$

Let $e^{2x} = t ⇒ 2e^{2x} dx = dt$.

When $x = \log \sqrt{2}, t = e^{2\log \sqrt{2}} = e^{\log 2} = 2$.

When $x = \log \sqrt{3}, t = e^{2\log \sqrt{3}} = e^{\log 3} = 3$.

$I = \frac{1}{2} \int\limits_{2}^{3} \frac{dt}{t(t - \frac{1}{t})} = \frac{1}{2} \int\limits_{2}^{3} \frac{dt}{t^2 - 1}$

$= \frac{1}{2} \left[ \frac{1}{2} \log \left| \frac{t-1}{t+1} \right| \right]_{2}^{3} = \frac{1}{4} \left( \log \frac{2}{4} - \log \frac{1}{3} \right) = \frac{1}{4} \log \frac{3}{2}$