Indefinite Integration
Integration of Rational Functions
Grade 12
Question:
<p>\(\int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}\,dx\) is equal to</p>
<p>(a) \(\log|e^{2x} - e^{-2x}| + C\)</p>
<p>(b) \(\frac{1}{2}\log|e^{2x} + e^{-2x}| + C\)</p>
<p>(c) Not given</p>
<p>(d) Not given</p>
Step-by-Step Solution
Key Concept: Recognize the integrand as a logarithmic derivative by identifying the numerator as related to the derivative of the denominator.
<p>Let \(u = e^{2x} + e^{-2x}\), then \(du = (2e^{2x} - 2e^{-2x})dx = 2(e^{2x} - e^{-2x})dx\). Therefore, \(\int \frac{e^{2x} - e^{-2x}}{e^{2x} + e^{-2x}}\,dx = \frac{1}{2}\int \frac{du}{u} = \frac{1}{2}\log|u| + C = \frac{1}{2}\log|e^{2x} + e^{-2x}| + C\).</p>
Correct Answer: B