Integrate [1/logx -1/(logx) ²]  dx

Let I = [1/logx -1/(logx) ²]  dx

Put log x=t (base of log, e^t=x) 
x=e^t 
dx =e^t dt

I= ∫ [1/t -1/t^2]e^t dt

Use formula [f(x) +f'(x) ]e^x=f(x). e^t + C
f'(x) =derivative of f(x) 

Integrate by partially (but f'(x) should not change because at last it is cancelled out. ) 

I=  ∫ (1/t) e^t dt  -∫ (1/t^2) e^t dt
I= I1 -I2

Integrate of I1 By ILATE RULE
  
I1=e^t/t +∫ [e^t/t^2)] dt

I= e^t/t +∫ [e^t/t^2)] dt - ∫ (1/t^2) e^t dt

I= e^t/t + C
 Put value of e^t and t in I equation

I = x/ log x + C
  

Answer of Integrate [1/logx -1/(logx) ²]  dx =  x/ log x + C

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