How do you integrate #-1 / (x(ln x)^2)#?

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Answer 1 · 2016-06-11T19:56:23

#1/lnx+C#

We have:

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

We can use substitution here, since the derivative of #lnx#, which is #1/x#, is present alongside #lnx#.

Let #u=lnx# such that #du=1/xdx#.

We then have:

#-int(lnx)^-2/xdx=-int(lnx)^-2(1/x)dx=-intu^-2du#

Integrate this with the rule: #intu^ndu=u^(n+1)/(n+1)+C#, where #n!=1#.

Thus,

#-intu^-2du=-(u^(-2+1)/(-2+1))+C=-(u^-1/(-1))+C=1/u+C#

Since #u=lnx#,

#=1/lnx+C#