The value of the integral \(\int_{\log_e 2}^{\log_e 3}\displaystyle\frac{e^{2x}-1}{e^{2x}+1}\,dx\) is:
Answer & explanation
Correct answer: option 2
The correct answer is Option (2) → \(\log_e 4-\log_e 3\)
Dividing the numerator and the denominator by \(e^x\):
\( \displaystyle \frac{e^{2x}-1}{e^{2x}+1} = \frac{e^{x}-e^{-x}}{e^{x}+e^{-x}} \)
The numerator is the derivative of the denominator, so the integral is a logarithm of the denominator:
\( \displaystyle \int_{\log_e 2}^{\log_e 3}\frac{e^{x}-e^{-x}}{e^{x}+e^{-x}}\,dx = \Big[\log_e\left(e^{x}+e^{-x}\right)\Big]_{\log_e 2}^{\log_e 3} \)
At the limits, \(e^{\log_e 3} = 3\) and \(e^{\log_e 2} = 2\):
\( \displaystyle = \log_e\left(3 + \frac{1}{3}\right) - \log_e\left(2 + \frac{1}{2}\right) \)
\( \displaystyle = \log_e\frac{10}{3} - \log_e\frac{5}{2} \)
\( \displaystyle = \log_e\left(\frac{10}{3}\times\frac{2}{5}\right) = \log_e\frac{4}{3} \)
\( = \log_e 4 - \log_e 3 \)