Question:
\(\int\frac{1}{e^{2x}-1}dx=\)
\(\int\frac{1}{e^{2x}-1}dx=\)
Updated On: Jun 10, 2024
- 2log|e2x-1|-x+C
- \(x-\frac{1}{2}\log|e2x-1|+C\)
- \(x+\frac{1}{2}\log|e2x-1|+C\)
- x-log|e2x-1|+C
- \(\frac{1}{2}\log|e2x-1|-x+C\)
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The Correct Option is
Solution and Explanation
The correct option is (E): \(\frac{1}{2}\log|e2x-1|-x+C\)
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