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Question

Mathematics Question on Integration

∫1e2x−1dx=\int\frac{1}{e^{2x}-1}dx=

A

2log|e2x-1|-x+C

B

x−12log⁡∣e2x−1∣+Cx-\frac{1}{2}\log|e2x-1|+C

C

x+12log⁡∣e2x−1∣+Cx+\frac{1}{2}\log|e2x-1|+C

D

x-log|e2x-1|+C

E

12log⁡∣e2x−1∣−x+C\frac{1}{2}\log|e2x-1|-x+C

Answer

12log⁡∣e2x−1∣−x+C\frac{1}{2}\log|e2x-1|-x+C

Explanation

Solution

The correct option is (E): 12log⁡∣e2x−1∣−x+C\frac{1}{2}\log|e2x-1|-x+C