How do you integrate #(ln x)^2/ x^3 dx#?

1 Answer
Apr 20, 2015

let's start by #u = ln(x)#
#du = 1/x#

#int ln^2(x)/x^3dx =int 1/x^2*1/x*ln^2(x) =int 1/x^2*u^2du #

#u = ln(x)#
#e^u=x#
#e^(2u) = x^2#
#e^(-2u)= 1/x^2#

So now we have :

#=>inte^(-2u)*u^2du#

By part :

#dv= e^(-2u)#
#v =-1/2e^(-2u)#

#w = u^2#
#dw = 2u#

#=>-1/2[u^2*e^(-2u)]+intu*e^(-2u)du#

By part again :

#dv = e^(-2u)#
#v = -1/2e^(-2u)#
#w = u#
#dw = 1#

#=>-1/2[u^2*e^(-2u)]-1/2[e^(-2u)*u]+1/2inte^(-2u)du#

#=>-1/2[u^2*e^(-2u)]-1/2[e^(-2u)*u]-1/4[e^(-2u)]#

Substitute back for #u = ln(x)#

#=>-1/2[ln^2(x)*1/x^2]-1/2[ln(x)*1/x^2]-1/4[1/x^2]+C#

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