 # the integral of ∫(1/lnx)dx is.......its 1 divided by natural log of x w.r.t. dx Badiuddin askIITians.ismu Expert
148 Points
13 years ago

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= et

dx =etdt

I =∫(1/t) etdt

=∫(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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2 years ago

et lnx =t

or x= et

dx =etdt

I =∫(1/t) etdt

=∫(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