Indefinite Integration
Exponential Functions
Grade 12
Question:
<p>Let \(I = \int \frac{e^{4x} - e^{2x}}{e^{4x} + e^{2x} + 1} dx\) and \(J = \int \frac{e^{2x}}{e^{4x} + e^{2x} + 1} dx\). Then, for an arbitrary constant C, the value of J – I equals [IIT - 2008]</p>
<p>(A) \(\frac{1}{2}\log\frac{e^{4x} - e^{2x} + 1}{e^{4x} + e^{2x} + 1} + C\)</p>
<p>(B) \(\frac{1}{2}\log\frac{e^{2x} - e^{x} + 1}{e^{2x} + e^{x} + 1} + C\)</p>
<p>(C) \(\frac{1}{2}\log\frac{e^{2x}}{e^{2x} + e^{x} + 1} + C\)</p>
<p>(D) \(\frac{1}{2}\log\frac{e^{4x} - e^{2x} + 1}{e^{4x} + e^{2x} + 1} + C\)</p>
Step-by-Step Solution
Key Concept: Find the difference J - I and recognize that the numerator is related to the derivative of the denominator.
<p>Calculate J - I = $\int \frac{e^{2x} - (e^{4x} - e^{2x})}{e^{4x} + e^{2x} + 1} dx = \int \frac{2e^{2x} - e^{4x}}{e^{4x} + e^{2x} + 1} dx$. Use substitution $u = e^{2x}$ and recognize the derivative of the denominator in the numerator to get the logarithmic form.</p>
Correct Answer: A