Mathematics

# $\displaystyle \int^{2}_{1} \dfrac{\log x}{x^{2}}.dx$

$\dfrac{-\log 2}{2}+\dfrac{1}{2}$

##### SOLUTION
$\int_1^2 {\dfrac{{\log x}}{{{x^2}}}dx}$
$\ = \int_1^2 {\dfrac{1}{x} \times \dfrac{{\log x}}{x}dx}$
putting $\log x = t \Rightarrow x = {e^t}$ and $\dfrac{1}{x}dx = dt$
$= \int_0^{\log 2} {\dfrac{t}{{{e^t}}}dt}$
$= \int_0^{\log 2} {t.{e^{ - t}}dt}$
$= \left[ { - t{e^{ - t}} + \int_{}^{} {{e^{ - t}}dt} } \right]$
$= \left[ { - t{e^{ - t}} - {e^{ - t}}} \right]_0^{\log 2}$
$= - \left[ {{e^{ - t}}\left( {t + 1} \right)} \right]_0^{\log 2}$
$= - \left[ {{e^{ - \log 2}}\left( {\log 2 + 1} \right) - 1} \right]$
$= - \left[ {\dfrac{1}{2}\left( {\log 2 + 1} \right) - 1} \right]$
$= - \left[ {\dfrac{1}{2}\log 2 - \dfrac{1}{2}} \right]$
$= \dfrac{{ - \log 2}}{2} + \dfrac{1}{2}$

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Single Correct Medium Published on 17th 09, 2020
Questions 203525
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