How do you integrate #int lnx/x^2# by parts?

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Answer 1 · 2017-07-30T19:43:09

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

#intuv'dx=uv-intu'vdx#

Here,

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

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

Therefore,

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