integration of 1/xlogx
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∫1/x.logx.dx
= ∫1/x. 1/logx .dx
Let logx = z { here logx with base e
logx = z
differentiate with respect to x
1/x = dz/dx
dx = x.dz ________(1)
put (1) in integration,
= ∫xdz/x.z
=∫dz/z
=logz { log z base e }
=log( logx ) + C
= ∫1/x. 1/logx .dx
Let logx = z { here logx with base e
logx = z
differentiate with respect to x
1/x = dz/dx
dx = x.dz ________(1)
put (1) in integration,
= ∫xdz/x.z
=∫dz/z
=logz { log z base e }
=log( logx ) + C
shal1:
its answer is log(logx)
whem logx base woth e then ots answer possibke
now see !! plz write clearly beclogx and lnx botu are fidferent thing oz l
here you wrote logx it means logx base 10 okay
you should write 1/x.lnx
okkk
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