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∫dx(ex−1)=\int \frac{dx}{(e^x - 1)} =

Solution

✅ Correct Option: 3

Multiply numerator and denominator by e−xe^{-x}: ∫e−x1−e−xdx\int \frac{e^{-x}}{1 - e^{-x}} dx. Let u=1−e−xu = 1 - e^{-x}, du=e−xdxdu = e^{-x} dx. Integral =log⁡∣1−e−x∣+C=log⁡∣ex−1ex∣+C= \log|1 - e^{-x}| + C = \log\left|\frac{e^x - 1}{e^x}\right| + C.

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