WolframAlpha gives small imaginary part for real integral $int_0^infty log(1 – 2xe^{-x} – e^{-2x}),dx$

Putting the integral $$int_0^infty log(1 – 2xe^{-x} – e^{-2x}),dx$$ into Wolfram|Alpha gives integral_0^∞ log(1 - e^(-2 x) - 2 e^(-x) x) dx = -6.89753 + 3.19343×10^-7 i

Two questions:

  1. Which numerical algorithm does it use that produces this small imaginary part?
  2. Is there a chance for a symbolic solution for this integral?