If ∫≤ft(1 + e^-x + e^-2x + …)dx = logφ(x) + C , then φ(x) is equal to:

The correct answer is Option (4) → \(e^x - 1\)

The integrand is an infinite geometric series with first term 1 and common ratio \(e^{-x}\). For x > 0, \(0 \lt e^{-x} \lt 1\), so its sum is

    \( \displaystyle 1 + e^{-x} + e^{-2x} + \ldots = \frac{1}{1 - e^{-x}} \)

    \( \displaystyle = \frac{e^x}{e^x - 1} \)

The numerator is the derivative of the denominator, so

    \( \displaystyle \int\frac{e^x}{e^x - 1}\,dx = \log\left(e^x - 1\right) + C \)

Comparing with \(\log\phi(x) + C\) gives \(\phi(x) = e^x - 1\).