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. \( \ln 3 - \ln 2 \)

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:

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