Mathematics

Solve :
$$\displaystyle \int \left (\log(\log x)+\dfrac{1}{(\log x)^2}\right)dx$$


SOLUTION
$$I=\displaystyle\int { \left( \log(\log x)+\dfrac { 1 }{ { (\log x) }^{ 2 } }  \right) dx } \\ \quad =\displaystyle\int { \log(\log x)dx } +\int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ Let\quad { I }_{ 1 }=\displaystyle\int { \log(\log x)dx } \\ \quad \quad \quad\quad =\displaystyle\int { \log(\log x).1dx } \\ \text{continue with integration by part}:\\ \displaystyle\int { f(x)g(x)dx } =f(x)\displaystyle\int { g(x)dx } -\int { \left[ f'(x)\int { g(x)dx }  \right]  } dx\\ now,\\ { { I }_{ 1 } }=\log(\log x)\displaystyle\int { 1 } dx\quad -\quad \int { \left[ \dfrac { d }{ dx } (\log(\log x)).\int { 1dx }  \right]  } dx\\ \quad \quad =x\log(\log x)\quad -\quad \displaystyle\int { \left[ \dfrac { 1 }{ \log x } .\dfrac { 1 }{ x } .x \right]  } dx\\ \quad \quad =x\log(\log x)\quad -\quad\displaystyle \int { \dfrac { 1 }{ \log x }  } dx\\ Let\quad { I }_{ 2 }=\displaystyle\int { \dfrac { 1 }{ \log x }  } dx\\ \quad \quad \quad \quad \quad =\displaystyle\int { \dfrac { 1 }{ \log x } .1dx } \\ \quad \quad \quad \quad \quad =\dfrac { 1 }{ \log x } .\displaystyle\int { 1dx } \quad -\quad \int { \left[ \dfrac { d }{ dx } \left( \dfrac { 1 }{ \log x }  \right) .\displaystyle\int { 1dx }  \right]  } dx\\ \quad \quad \quad \quad \quad =\quad \dfrac { x }{ \log x } \quad -\quad \displaystyle\int { \dfrac { -1 }{ { (\log x) }^{ 2 } } .\dfrac { 1 }{ x } .xdx } \\ \quad \quad \quad \quad \quad =\dfrac { x }{ \log x } +\displaystyle\int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ putting\quad the\quad value\quad of\quad { I }_{ 2 }\quad in\quad { I }_{ 1 }\\ { I }_{ 1 }=x\log(\log x)\quad -\quad \dfrac { x }{ \log x } \quad -\quad \displaystyle\int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ now\quad putting\quad the\quad value\quad of\quad { I }_{ 1 }\quad in\quad I\\ I\quad =\quad x\log(\log x)\quad -\quad \dfrac { x }{ \log x } \quad -\quad\displaystyle \int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \quad +\quad \int { \dfrac { 1 }{ { (\log x) }^{ 2 } } dx } \\ \quad \quad =\quad x\log(\log x)\quad -\quad \dfrac { x }{ \log x } \quad +\quad C,\quad where\quad C\quad =\quad integration\quad constant\quad$$
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Subjective Medium Published on 17th 09, 2020
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