Mathematics

The value of $\displaystyle\int {\dfrac{{\ln n\left( {1 - \left({\dfrac{1}{x}} \right)} \right)dx}}{{x\left( {x - 1} \right)}}}$ is

$\dfrac{1}{2}\left[ {m{{\left( {1 - \frac{1}{x}} \right)}^2}} \right] + C$

SOLUTION

Consider, $\displaystyle I=\int {{{ln\left( {1 - {1 \over x}} \right)} \over {x\left( {x - 1} \right)}}} dx$

$\displaystyle I = \int {{{\ln\left( {1 - {1 \over x}} \right)} \over {{x^2}\left( {1 - {1 \over x}} \right)}}} dx$

Put $\ln\left( {1 - {\dfrac 1 x}} \right) = t$ $\Rightarrow$ $\displaystyle {1 \over {\left( {1 - {\dfrac 1 x}} \right)}} \times {\dfrac {1} {{x^2}}}dx = dt$

$\displaystyle I= \int {tdt}$

$\displaystyle I= {{{t^2}} \over 2} + c$

$\displaystyle I= {1 \over 2}{\left[ {ln\left( {1 - {1 \over x}} \right)} \right]^2} + c$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
Enrolled Students 86

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