the integral of ∫(1/lnx)dx is.......its 1 divided by natural log of x w.r.t. dx

Dear kshitij

you can find the solution of these type of integral in terms of infinite series

I=∫(1/lnx)dx

let lnx =t

     or x= e^t

        dx =e^tdt

I =∫(1/t) e^tdt

  =∫(1/t)[1+t+t²/2! + t³/3! +..........]dt

  =∫[1/t +1+t/2! + t²/3! +..........]dt

  =lnt + t + t²/2*2! + t³/3* 3! +.........

  =ln(lnx) + lnx + (lnx)²/2*2! + (lnx)³/3* 3! +.........

  =ln(lnx) +  ∑(lnx)^r/r*r!  where r varies from 1 to infinity

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