∫1ln x−1(ln x)2dx=

I=∫1lnx-1(lnx)2dx put lnx=t x=et dx=etdt I=∫1t-1t2et dt =1t.et+c =1lnx.x+c

∫1ln x−1(ln x)2dx=

I=∫1lnx-1(lnx)2dx put lnx=t x=et dx=etdt I=∫1t-1t2et dt =1t.et+c =1lnx.x+c

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$\int (\frac{1}{\ln x} - \frac{1}{{(\ln x)}^{2}}) d x =$

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$\frac{- e^{x}}{{(\ln x)}^{2}} + c$

$e^{x} (\frac{1}{\ln x}) + c$

$\frac{- x}{{(\ln x)}^{2}} + c$

$\frac{x}{\ln x} + c$

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Detailed Solution

$I = \int (\frac{1}{lnx} - \frac{1}{{(lnx)}^{2}}) dx put lnx = t x = e^{t} dx = e^{t} dt I = \int (\frac{1}{t} - \frac{1}{t^{2}}) e^{t} dt = \frac{1}{t} . e^{t} + c = (\frac{1}{lnx} . x) + c$

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∫1ln x−1(ln x)2dx=

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$\int (\frac{1}{\ln x} - \frac{1}{{(\ln x)}^{2}}) d x =$