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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)$

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$f(x)=x^6+2x^4+x^3+2x+3 $

$\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$

$n=?$

Let $f(x)=x^6+2x^4+x^3+2x+3,x \in R $. Then the natural number n for which $\mathop {\lim }\limits_{x \to 1} \frac{{{x^n}f(1) - f(x)}}{{x - 1}} = 44$ is _ _ _ _ . Solution Since the limit has $\left[ {\frac{0}{0}} \right]$ form, L.H. Rule is applicable. Thus, $\mathop {\lim }\limits_{x \to 1} n{x^{n - 1}}f(1) - f'(x) = 44$ $\therefore nf(1) - f'(1) = 44$ $\therefore n.9 - ({6.1^5} + {8.1^3} + {3.1^2} + 2.1) = 44$ $ \Rightarrow 9n - 19 = 44$ $\Rightarrow n=7$