Question #5266a

1 Answer
Nov 9, 2016

#-1/(3(ln(x))^3)+C#

Explanation:

#I=int1/(x(ln(x))^4)dx#

We can use substitution #u=ln(x)#. Differentiating this shows that #(du)/dx=1/x#, or #du=1/xdx#. This is already present in our integral, but we can make it clearer:

#I=int1/(ln(x))^4(1/xdx)=intunderbrace((ln(x))^-4)_(u^-4)overbrace((1/xdx))^(du)=intu^-4du#

We can now use the integration rule #intu^ndu=u^(n+1)/(n+1)+C#:

#I=u^(-4+1)/(-4+1)+C=u^(-3)/(-3)+C=-1/(3u^3)+C#

Since #u=ln(x)#:

#I=-1/(3(ln(x))^3)+C#