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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= 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
Please feel free to post as many doubts on our discussion forum as you can.If you find any question Difficult to understand - post it here and we will get you the answer and detailed solution very quickly. We are all IITians and here to help you in your IIT JEE preparation.All the best. Regards,Askiitians ExpertsBadiuddin
et lnx =t
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