The given integral =$ – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{x(1 + x)}}dx} $

$ = \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{1 + x}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = \int\limits_{ – 1}^0 {\frac{{\ln \left[ {1 + ( – 1 – x) + {{( – 1 – x)}^2}} \right]}}{{1 + (-1-x)}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{{ – x}}dx} – \int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = – 2\int\limits_{ – 1}^0 {\frac{{\ln (1 + x + {x^2})}}{x}dx} $

$ = – 2\int\limits_{ – 1}^0 {\frac{{\ln \frac{{(1 – x)(1 + x + {x^2})}}{{(1 – x)}}}}{x}dx} $

$ = 2\int\limits_{ – 1}^0 {\frac{{\ln \frac{{(1 – x)}}{{(1 – {x^3})}}}}{x}dx} $

$ = 2\int\limits_{ – 1}^0 {\frac{{\ln (1 – x)}}{x}dx – 2\int\limits_{ – 1}^0 {\frac{{\ln (1 – {x^3})}}{x}dx} } $ ……(*)

Consider, $I = \int\limits_{ – 1}^0 {\frac{{\ln (1 – {x^3})}}{x}dx} $

Let, ${x^3} = t$

$ \Rightarrow 3{x^2}dx = dt$

$ \Rightarrow \frac{{dx}}{x} = \frac{{dt}}{{3{x^3}}} = \frac{{dt}}{{3t}}$

So, $I = \int\limits_{ – 1}^0 {\frac{{\ln (1 – t)}}{{3t}}dt} $

I can also be written as $\int\limits_{ – 1}^0 {\frac{{\ln (1 – x)}}{{3x}}dx} $

From (*), the given integral

$ = 2\int\limits_{ – 1}^0 {\frac{{\ln (1 – x)}}{x}dx – 2\int\limits_{ – 1}^0 {\frac{{\ln (1 – x)}}{{3x}}dx} } $

$ = \frac{4}{3}\int\limits_{ – 1}^0 {\frac{{\ln (1 – x)}}{x}dx} $

$ = \frac{4}{3}\int\limits_{ – 1}^0 {\frac{{ – x – \frac{{{x^2}}}{2} – \frac{{{x^3}}}{3} – \frac{{{x^4}}}{4} – ……..}}{x}dx} $

$ = – \frac{4}{3}\int\limits_{ – 1}^0 {1 + \frac{x}{2} + \frac{{{x^2}}}{3} + \frac{{{x^3}}}{4} + ………dx} $

$ = – \frac{4}{3}\left. {\left( {x + \frac{{{x^2}}}{{{2^2}}} + \frac{{{x^3}}}{{{3^2}}} + \frac{{{x^4}}}{{{4^2}}} + ……….} \right)} \right|_{ – 1}^0$

$ = – \frac{4}{3}\left[ {0 – \left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ………} \right)} \right]$

$ = \frac{4}{3}\left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ………} \right)$