How do you find the integral of #int ln x/(x^3) dx# from 1 to infinity?

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Answer 1 · 2018-03-26T05:36:52

The answer is #=1/4#

First , calculate the indefinite integral

#I=int(lnxdx)/(x^3)#

#intuv'=uv-intu'v#

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

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

Therefore,

#I=-lnx/(2x^2)-int(-(1dx)/(2x^3))#

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

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

Now, calculate the definite intergal

#int_1^t(lnxdx)/(x^3)=[-lnx/(2x^2)-1/(4x^2)]_1^t#

#=(-lnt/(2t^2)-1/(4t^2))-(-ln1/(2*1^2)-1/(4*1^2))#

#=1/4-lnt/(2t^2)-1/(4t^2)#

#lim_(t->oo)-lnt/(2t^2)=0#

#lim_(t->oo)-1/(4t^2)=0#

And finally the integral is

#int_1^ oo(lnxdx)/(x^3)= lim_(t->oo)(1/4-lnt/(2t^2)-1/(4t^2))#

#=1/4#