The value of the integral $\int\limits_{\log _e 2}^{\log _e 3} \frac{e^{2 x}-1}{e^{2 x}+1} d x$ is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → $\log _e 4 - \log _e 3$
$\int\limits_{\log_2}^{\log_3} \frac{e^{2 x}-1}{e^{2 x}+1} d x$
$=\int\limits_{\log_2}^{\log_3}\frac{e^x-e^{-x}}{e^x+e^{-x}}dx$
$=\left|\log(e^x+e^{-x})\right|_{\log_2}^{\log_3}$
$=\log(3+\frac{1}{3})=\log(2+\frac{1}{2})$
$=\log(\frac{10}{3})-\log(\frac{5}{2})$
$=\log(2)+\log(5)-\log(3)-\log(5)+\log(2)$
$=\log(4)-\log(3)$