Mathematics

# Solve: $\int {\dfrac{{{\mathop{\rm logx}\nolimits} }}{{{{\left( {1 + \log x} \right)}^2}}}dx}$

##### SOLUTION
$\begin{array}{l}\int {\cfrac{{{\mathop{\rm logx}\nolimits} }}{{{{\left( {1 + \log x} \right)}^2}}}dx} \\Let,\log x = t\\x = {e^t}\\dx = xdt\\\therefore I = \int {\cfrac{t}{{{{\left( {1 + t} \right)}^2}}}xdt} = \int {\cfrac{{{e^t}.t}}{{{{\left( {1 + t} \right)}^2}}}dt} = \int {\cfrac{{{e^t}.\left( {t + 1 - 1} \right)}}{{{{\left( {1 + t} \right)}^2}}}dt} \\ = \int {\cfrac{{{e^t}.\left( {t + 1} \right)}}{{{{\left( {1 + t} \right)}^2}}}dt} - \int {\cfrac{{{e^t}}}{{\left( {1 + {t^2}} \right)}}dt} \\ = \int {{e^t}\left[ {\cfrac{1}{{1 + t}} - \cfrac{1}{{1 + {t^2}}}} \right]} dx\\ = \int {{e^t}\left[ {f(x) + f'(x)} \right]} x\\ = {e^t}f(x) + C\\ = {e^t}\cfrac{1}{{1 + t}} + C\\ = \cfrac{x}{{1 + \log x}} + C\end{array}$

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Subjective Medium Published on 17th 09, 2020
Questions 203525
Subjects 9
Chapters 126
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