Mathematics

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


ANSWER

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


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
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