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.
Let:
I = \(\int_{\log_{e} 2}^{\log_{e} 3} \frac{e^{2x} - 1}{e^{2x} + 1} dx\).
Perform the substitution \( u = e^{2x} \). This yields \( du = 2e^{2x}dx \), which can be rewritten as \( dx = \frac{du}{2u} \). The new limits of integration are:
\( x = \log_{e} 2 \implies u = e^{2 \cdot \log_{e} 2} = 4 \),
\( x = \log_{e} 3 \implies u = e^{2 \cdot \log_{e} 3} = 9 \).
The integral transforms to:
I = \(\frac{1}{2} \int_{4}^{9} \frac{u - 1}{u(u + 1)} du\).
Decompose the integrand using partial fractions:
\(\frac{u - 1}{u(u + 1)} = \frac{1}{u} - \frac{1}{u + 1}\).
Therefore:
I = \(\frac{1}{2} \int_{4}^{9} \left( \frac{1}{u} - \frac{1}{u + 1} \right) du\).
Evaluate the integral:
I = \(\frac{1}{2} [\ln u - \ln(u + 1)]_{4}^{9}\).
Substitute the limits and simplify:
I = \(\frac{1}{2} \left[(\ln 9 - \ln 10) - (\ln 4 - \ln 5)\right]\).
Combine logarithmic terms:
I = \(\frac{1}{2} \left[\ln \left(\frac{9}{10}\right) - \ln \left(\frac{4}{5}\right)\right]\).
Further simplification yields:
I = \(\frac{1}{2} \ln \left(\frac{9}{10} \times \frac{5}{4}\right)\).
Simplify the expression inside the logarithm:
I = \(\frac{1}{2} \ln \left(\frac{45}{40}\right)\).
This simplifies to:
I = \(\ln 4 - \ln 3\).
Which is equivalent to:
\(\ln_{e} 4 - \ln_{e} 3\).