Evaluating the Definite Integral

We are asked to find the value of the definite integral \( \int_{\log_e 2}^{\log_e 3} \frac{e^{2x}- 1}{e^{2x} + 1} dx \).

Simplifying the Integrand using Hyperbolic Functions

Let's simplify the integrand \( \frac{e^{2x}- 1}{e^{2x} + 1} \). We can divide both the numerator and the denominator by \(e^x\):

\( \frac{e^{2x}- 1}{e^{2x} + 1} = \frac{e^x(e^x - e^{-x})}{e^x(e^x + e^{-x})} = \frac{e^x - e^{-x}}{e^x + e^{-x}} \)

This simplified expression is the definition of the hyperbolic tangent function, \( \tanh(x) \). So, the integrand is \( \tanh(x) \).

The integral can be rewritten as \( \int_{\log_e 2}^{\log_e 3} \tanh(x) dx \).

Calculating the Integral of tanh(x)

The standard integral of \( \tanh(x) \) is given by:

\( \int \tanh(x) dx = \log|\cosh(x)| + C \)

where \( \cosh(x) = \frac{e^x + e^{-x}}{2} \).

Since \( \cosh(x) \) is always positive for real \( x \), we can write the integral as \( \log(\cosh(x)) + C \).

Evaluating the Definite Integral at the Limits

Now, we evaluate the definite integral \( [\log(\cosh(x))]_{\log_e 2}^{\log_e 3} \).

First, calculate \( \cosh(x) \) at the upper and lower limits:

• At the upper limit \(x = \log_e 3\):
  \( \cosh(\log_e 3) = \frac{e^{\log_e 3} + e^{-\log_e 3}}{2} \)
• Using the property \( e^{\log_b a} = a \) and \( e^{-\log_b a} = e^{\log_b a^{-1}} = a^{-1} \), we get:
  \( e^{\log_e 3} = 3 \) and \( e^{-\log_e 3} = 3^{-1} = \frac{1}{3} \)
• So, \( \cosh(\log_e 3) = \frac{3 + 1/3}{2} = \frac{9/3 + 1/3}{2} = \frac{10/3}{2} = \frac{10}{6} = \frac{5}{3} \).
• At the lower limit \(x = \log_e 2\):
  \( \cosh(\log_e 2) = \frac{e^{\log_e 2} + e^{-\log_e 2}}{2} \)
• Using the same properties:
  \( e^{\log_e 2} = 2 \) and \( e^{-\log_e 2} = 2^{-1} = \frac{1}{2} \)
• So, \( \cosh(\log_e 2) = \frac{2 + 1/2}{2} = \frac{4/2 + 1/2}{2} = \frac{5/2}{2} = \frac{5}{4} \).

Now, substitute these values into the definite integral formula \( F(b) - F(a) \):

\( [\log(\cosh(x))]_{\log_e 2}^{\log_e 3} = \log\left(\cosh(\log_e 3)\right) - \log\left(\cosh(\log_e 2)\right) \)

\( = \log\left(\frac{5}{3}\right) - \log\left(\frac{5}{4}\right) \)

Using Logarithm Properties to Simplify

We use the logarithm property \( \log_b a - \log_b c = \log_b\left(\frac{a}{c}\right) \):

\( \log\left(\frac{5}{3}\right) - \log\left(\frac{5}{4}\right) = \log\left(\frac{5/3}{5/4}\right) \)

\( = \log\left(\frac{5}{3} \times \frac{4}{5}\right) \)

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

The value of the integral is \( \log_e\left(\frac{4}{3}\right) \).

Comparing with Options

Let's express the given options in the form \( \log_e(\text{single number}) \):

• Option 1: \( \log_e 3 \)
• Option 2: \( \log_e 3 - \log_e 2 = \log_e\left(\frac{3}{2}\right) \)
• Option 3: \( \log_e 9 - \log_e 4 = \log_e\left(\frac{9}{4}\right) \)
• Option 4: \( \log_e 4 - \log_e 3 = \log_e\left(\frac{4}{3}\right) \)

Our calculated value \( \log_e\left(\frac{4}{3}\right) \) matches the value in Option 4.

Revision Table: Integral Evaluation & Logarithms

Additional Information on Calculus and Hyperbolic Functions

This problem demonstrates the power of recognizing functions. The integrand, though initially looking complicated, simplifies nicely into a standard hyperbolic function, \( \tanh(x) \). Knowing the integral of \( \tanh(x) \) is crucial for a quick solution.

The definition of \( \tanh(x) \) comes from the ratio of hyperbolic sine (\( \sinh(x) = \frac{e^x - e^{-x}}{2} \)) and hyperbolic cosine (\( \cosh(x) \)). Just like trigonometric functions relate to a circle, hyperbolic functions relate to a hyperbola.

Definite integrals have many applications, such as finding areas, volumes, displacement, and work done. The limits of integration are essential as they define the specific interval of interest.

Logarithm properties are frequently used when evaluating definite integrals, especially when the antiderivative involves logarithmic or inverse trigonometric/hyperbolic functions. Remembering properties like \( \log a - \log b = \log(a/b) \) helps simplify the final result.