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$I = \int\limits_0^1 {\frac{{\ln (1 + x + {x^2} + .........\infty \, terms)}}{x}dx} $

A) $I = 1 + \frac{1}{2} + \frac{1}{3} + \frac{1}{4} + ........\infty \, terms$

B) $I = 1 + \frac{1}{{2!}} + \frac{1}{{3!}} + \frac{1}{{4!}} + ........\infty \, terms$

C) $I = 1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + ........\infty \, terms$

D) $I = 1 - \frac{1}{2} + \frac{1}{3} - \frac{1}{4} + ........\infty \, terms$

Since x lies between 0 and 1, the infinite terms summation formula of an infinite G.P. with x as common ratio can be used here.

$I = \int\limits_0^1 {\frac{{\ln \left( {\frac{1}{{1 - x}}} \right)}}{x}dx} $

$ = \int\limits_0^1 {\frac{{ - \ln (1 - x)}}{x}dx} $

$ = \int\limits_0^1 {\frac{{x + \frac{{{x^2}}}{2} + \frac{{{x^3}}}{3} + \frac{{{x^4}}}{4} + .......}}{x}dx} $

$ = \int\limits_0^1 {1 + \frac{x}{2} + \frac{{{x^2}}}{3} + \frac{{{x^3}}}{4} + .......dx} $

$ = \left. {x + \frac{{{x^2}}}{{{2^2}}} + \frac{{{x^3}}}{{{3^2}}} + \frac{{{x^4}}}{{{4^2}}} + .........} \right|_0^1$

$ = 1 + \frac{1}{{{2^2}}} + \frac{1}{{{3^2}}} + \frac{1}{{{4^2}}} + .........$

Hence, option (C).

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