The value of the integral

Correct Answer: log_e 4 − log_e 3

Step-by-step Explanation:

Based on the provided image, the definite integral to evaluate is:

$$I={\int }_{{\log }_{e}2}^{{\log }_{e}3}\frac{e^{2x}-1}{e^{2x}+1}dx$$

Step 1: Simplify the integrand
We can divide both the numerator and the denominator of the integrand by $e^x$:

$$I={\int }_{{\log }_{e}2}^{{\log }_{e}3}\frac{e^x-e^{-x}}{e^x+e^{-x}}dx$$

Step 2: Substitution Method
Let us use the substitution method by setting the denominator as $u$:

$$u=e^x+e^{-x}$$

Differentiating both sides with respect to $x$, we obtain:

$$du=(e^x-e^{-x})dx$$

Step 3: Change the limits of integration
We must update the limits of integration from $x$ to $u$:
• For the lower limit, when $x={\log }_{e}2$:

$$u=e^{{\log }_{e}2}+e^{-{\log }_{e}2}=2+\frac{1}{2}=\frac{5}{2}$$

• For the upper limit, when $x={\log }_{e}3$:

$$u=e^{{\log }_{e}3}+e^{-{\log }_{e}3}=3+\frac{1}{3}=\frac{10}{3}$$

Step 4: Integrate with respect to $u$
Substitute these components back into the integral:

$$I={\int }_{5/2}^{10/3}\frac{1}{u}du$$

Evaluating the integral gives:

$$I={\left[{\log }_{e}u\right]}_{5/2}^{10/3}$$

Applying the limits:

$$I={\log }_e\left(\frac{10}{3}\right)-{\log }_e\left(\frac{5}{2}\right)$$

Step 5: Apply Logarithmic Properties
Using the quotient property of logarithms, ${\log }_{e}A-{\log }_{e}B={\log }_e\left(\frac{A}{B}\right)$:

$$I={\log }_e\left(\frac{10/3}{5/2}\right)$$

$$I={\log }_e(\frac{10}{3}\times \frac{2}{5})$$

$$I={\log }_e\left(\frac{4}{3}\right)$$

Expanding this back with logarithmic subtraction:

$$I={\log }_{e}4-{\log }_{e}3$$