Mathematics

Find the antiderivative of $$f\left( x \right) = \,ln\,\left( {ln\,x} \right) + {\left( {ln\,x} \right)^{ - 2}}$$ whose graph passes through $$ (e, e)$$.


SOLUTION
$$\begin{array}{l} We\, get \\ f\left( x \right) =\ln { \left( { \ln { x }  } \right)  } +\frac { 1 }{ { { { \left( { \ln { x }  } \right)  }^{ 2 } } } }  \\ \int { f(x)dx } =\int { \left( { \ln { \left( { \ln { x }  } \right)  } +\frac { 1 }{ { { { \left( { \ln { x }  } \right)  }^{ 2 } } } }  } \right)  } dx \\ x={ e^{ t } }\, \, ,dx={ e^{ t } }dt \\ =\int { \left( { \ln { t } +\frac { 1 }{ { { t^{ 2 } } } }  } \right)  } { e^{ t } }dt \\ =\int { { e^{ t } }\left( { { { lnt } }+\frac { 1 }{ t }  } \right) dt } +\int { { e^{ t } }\left( { \frac { 1 }{ { { t^{ 2 } } } } -\frac { 1 }{ t }  } \right) dt }  \\ =\int { { e^{ t } }\left( { \ln { t } +\frac { { d\left( { \ln { t }  } \right)  } }{ { dt } }  } \right) dt+\int { { e^{ t } }\left( { -\frac { 1 }{ t } +\frac { d }{ { dt } } \left( { \frac { { -1 } }{ t }  } \right) dt } \right)  }  }  \\ ={ e^{ t } }\ln { t } -\frac { { { e^{ t } } } }{ t } +C \\ \int { fdx } =x\ln { \left( { \ln { x }  } \right)  } -\frac { x }{ { \ln { x }  } } +C \\ y=x\ln { \left( { \ln { x }  } \right)  } \frac { { -e } }{ 1 } +c\, \, put\, \, \left( { e,e } \right)  \\ e=e\ln { \left( 1 \right)  } -\frac { e }{ 1 } +C\, \, \, \,  \\ c=2e \\ y=x\ln { \left( { { { lnx } } } \right)  } \frac { { -x } }{ { \ln { x }  } } +2e \end{array}$$

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Subjective Medium Published on 17th 09, 2020
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