Prove that, $\int\limits_0^1 {\frac{1}{{x(1 + x)}}.\ln \left[ {\frac{1}{{1 + x(1 + x)}}} \right]dx} =$$\frac{4}{3}\left( {1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ........\infty \, terms} \right)$

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

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

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

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

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

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

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

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