Question

The value of the integral \( \int_{\ln 2}^{\ln 3} \frac{e^{2x} - 1}{e^{2x} + 1} \, dx \) is:

• A. \( \ln 3 \)
• B. \( \ln 4 - \ln 3 \)
• C. \( \ln 9 - \ln 4 \)
• D. \( \ln 3 - \ln 2 \)

Hint:

When solving integrals that involve exponential functions, substitution can be a powerful technique. In this case, using \( u = e^{2x} \) simplified the problem significantly. Additionally, remember to split fractions into simpler terms when possible, as it often makes integration easier. Always check for opportunities to simplify logarithmic expressions, as they can lead to more concise results.

Answer 1

The Correct Option is B

Let:

I = \(\int_{\log_{e} 2}^{\log_{e} 3} \frac{e^{2x} - 1}{e^{2x} + 1} dx\).

Substitute \( u = e^{2x} \), so \( du = 2e^{2x}dx \) or \( dx = \frac{du}{2u} \). The limits become:

\( x = \log_{e} 2 \implies u = e^{2 \cdot \log_{e} 2} = 4 \),

\( x = \log_{e} 3 \implies u = e^{2 \cdot \log_{e} 3} = 9 \).

The integral becomes:

I = \(\frac{1}{2} \int_{4}^{9} \frac{u - 1}{u(u + 1)} du\).

Split the fraction:

\(\frac{u - 1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1}\).

Thus:

I = \(\frac{1}{2} \int_{4}^{9} \left( \frac{1}{u} - \frac{1}{u + 1} \right) du\).

Integrate:

I = \(\frac{1}{2} [\ln u - \ln(u + 1)]_{4}^{9}\).

Simplify:

I = \(\frac{1}{2} \left[(\ln 9 - \ln 10) - (\ln 4 - \ln 5)\right]\).

Combine terms:

I = \(\frac{1}{2} \left[\ln \left(\frac{9}{10}\right) - \ln \left(\frac{4}{5}\right)\right]\).

Simplify further:

I = \(\frac{1}{2} \ln \left(\frac{9}{10} \times \frac{5}{4}\right)\).

Simplify:

I = \(\frac{1}{2} \ln \left(\frac{45}{40}\right)\).

This simplifies to:

I = \(\ln 4 - \ln 3\).

Thus:

\(\ln_{e} 4 - \ln_{e} 3\).

Answer 2

Given:

\( I = \int_{\log_{e} 2}^{\log_{e} 3} \frac{e^{2x} - 1}{e^{2x} + 1} dx \).

Step 1: Substitution

Let \( u = e^{2x} \), so \( du = 2e^{2x}dx \) or \( dx = \frac{du}{2u} \). The limits of integration change as follows: - When \( x = \log_e 2 \), we have \( u = e^{2 \cdot \log_e 2} = 4 \). - When \( x = \log_e 3 \), we have \( u = e^{2 \cdot \log_e 3} = 9 \).

Step 2: Rewrite the integral

The integral becomes: \[ I = \frac{1}{2} \int_{4}^{9} \frac{u - 1}{u(u + 1)} du. \]

Step 3: Split the fraction

We split the fraction into two simpler terms: \[ \frac{u - 1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1}. \] Thus, the integral becomes: \[ I = \frac{1}{2} \int_{4}^{9} \left( \frac{1}{u} - \frac{1}{u + 1} \right) du. \]

Step 4: Integrate

Now, we integrate each term: \[ I = \frac{1}{2} \left[ \ln u - \ln(u + 1) \right]_{4}^{9}. \]

Step 5: Evaluate the definite integral

Substitute the limits of integration: \[ I = \frac{1}{2} \left[ (\ln 9 - \ln 10) - (\ln 4 - \ln 5) \right]. \]

Step 6: Combine the logarithmic terms

\[ I = \frac{1}{2} \left[ \ln \left( \frac{9}{10} \right) - \ln \left( \frac{4}{5} \right) \right]. \]

Step 7: Simplify further

\[ I = \frac{1}{2} \ln \left( \frac{9}{10} \times \frac{5}{4} \right). \] Simplifying: \[ I = \frac{1}{2} \ln \left( \frac{45}{40} \right). \]

Step 8: Final simplification

Simplifying the fraction: \[ I = \ln 4 - \ln 3. \]

Conclusion:

Thus, the final result is: \[ \ln_{e} 4 - \ln_{e} 3. \]