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If ∫(1+e−x+e−2x+...)dx=log⁡ϕ(x)+C\int (1 + e^{-x} + e^{-2x} + ...)dx = \log\phi(x) + C, then ϕ(x)\phi(x) is equal to :

Solution

✅ Correct Option: 4

The series is GP with first term 11 and ratio e−xe^{-x}: sum =11−e−x=exex−1= \frac{1}{1 - e^{-x}} = \frac{e^x}{e^x - 1}. Integrating: ∫exex−1dx=log⁡∣ex−1∣+C\int \frac{e^x}{e^x - 1} dx = \log|e^x - 1| + C. So ϕ(x)=ex−1\phi(x) = e^x - 1.

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