Skip to main content

Visit this link for 1 : 1 LIVE Classes.

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)$

Popular posts from this blog

${\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$

The range of the function $f(x) = {\log _{\sqrt 5 }}\left[ {3 + \cos \left( {\frac{{3\pi }}{4} + x} \right) + \cos \left( {\frac{\pi }{4} + x} \right) + \cos \left( {\frac{\pi }{4} - x} \right) - \cos \left( {\frac{{3\pi }}{4} - x} \right)} \right]$ is: (A) $[ - 2,2]$ (B) $\left[ {\frac{1}{{\sqrt 5 }},\sqrt 5 } \right]$ (C) $(0,\sqrt 5 )$ (D) $[ 0,2]$ Solution We have, $f(x) = {\log _{\sqrt 5 }}\left( {3 - 2\sin \frac{{3\pi }}{4}\sin x + 2\cos \frac{\pi }{4}\cos x} \right)$ $ \Rightarrow f(x) = {\log _{\sqrt 5 }}\left[ {3 + \sqrt 2 (\cos x - \sin x)} \right]$ Now, $ - \sqrt 2  \le \cos x - \sin x \le \sqrt 2 $ $\therefore - 2 \le \sqrt 2 (\cos x - \sin x) \le 2$ $\therefore 1 \le 3 + \sqrt 2 (\cos x - \sin x) \le 5$ $\therefore{\log _{\sqrt 5 }}1 \le {\log _{\sqrt 5 }}[3 + \sqrt 2 (\cos x - \sin x)] \le {\log _{\sqrt 5 }}5$ $ \Rightarrow 0 \le f(x) \le 2$ Answer: (D)
Read more