How do solve #int lnx/x^3 dx# ?

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Answer 1 · 2016-12-24T07:41:29

The answer is #=-(2lnx+1)/(4x^2)+C#

We need

#intx^ndx=x^(n+1)/(n+1)+C(n!=-1)#

#(lnx)'=1/x#

#intuv'dx=uv-intu'vdx#

Here,

#u=lnx, #=>, #u'=1/x#

#v'=1/x^3#, #=>#, #v=-1/(2x^2)#

Therefore,

#int(lnxdx)/x^3=-lnx/(2x^2)+1/2intdx/x^3#

#=-lnx/(2x^2)-1/(4x^2)+C#

#=-(2lnx+1)/(4x^2)+C#