# 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?