The value of the integral $\int_{\log _{e} 2}^{\log _{e} 3} \frac{e^{2 x}-1}{e^{2 x}+1} d x$ is :

Simplify:

$\dfrac{e^{2x}-1}{e^{2x}+1} = \dfrac{e^{2x}+1-2}{e^{2x}+1} = 1 - \dfrac{2}{e^{2x}+1}$

So our integral becomes:

$\int_{\log_e 2}^{\log_e 3} \left(1 - \dfrac{2}{e^{2x}+1}\right) dx = \int_{\log_e 2}^{\log_e 3} dx - 2\int_{\log_e 2}^{\log_e 3} \dfrac{1}{e^{2x}+1} dx$

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The first part is straightforward:

$\int_{\log_e 2}^{\log_e 3} dx = x\big|_{\log_e 2}^{\log_e 3} = \log_e 3 - \log_e 2 = \log_e \dfrac{3}{2}$

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For the second part, we use substitution $u = e^{2x}$, giving:

$du = 2e^{2x}dx$ and $dx = \dfrac{du}{2u}$

When $x = \log_e 2$, $u = e^{2\log_e 2} = 4$

When $x = \log_e 3$, $u = e^{2\log_e 3} = 9$

The integral becomes:

$\int_4^9 \dfrac{1}{u+1} \cdot \dfrac{du}{2u} = \dfrac{1}{2}\int_4^9 \dfrac{1}{u(u+1)} du$

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Using partial fractions:

$\dfrac{1}{u(u+1)} = \dfrac{1}{u} - \dfrac{1}{u+1}$

Therefore:

$\dfrac{1}{2}\int_4^9 \dfrac{1}{u(u+1)} du = \dfrac{1}{2}\int_4^9 \left(\dfrac{1}{u} - \dfrac{1}{u+1}\right) du$

$= \dfrac{1}{2}\left[\ln|u| - \ln|u+1|\right]_4^9$

$= \dfrac{1}{2}\left[\ln\left|\dfrac{u}{u+1}\right|\right]_4^9$

$= \dfrac{1}{2}\left[\ln\left|\dfrac{9}{10}\right| - \ln\left|\dfrac{4}{5}\right|\right]$

$= \dfrac{1}{2}\ln\left|\dfrac{9/10}{4/5}\right| = \dfrac{1}{2}\ln\left|\dfrac{9 \cdot 5}{10 \cdot 4}\right| = \dfrac{1}{2}\ln\left|\dfrac{45}{40}\right| = \dfrac{1}{2}\ln\left|\dfrac{9}{8}\right|$

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Combining the results:

$\int_{\log_e 2}^{\log_e 3} \dfrac{e^{2x}-1}{e^{2x}+1} dx = \log_e \dfrac{3}{2} - \ln\dfrac{9}{8}$

$= \ln\dfrac{3}{2} - \ln\dfrac{9}{8} = \ln\left(\dfrac{3/2}{9/8}\right) = \ln\left(\dfrac{3 \cdot 8}{2 \cdot 9}\right) = \ln\left(\dfrac{24}{18}\right) = \ln\left(\dfrac{4}{3}\right)$

The value of the integral is $\ln\left(\dfrac{4}{3}\right) = \log _{e} 4-\log _{e} 3$