What is the antiderivative of #(ln x) ^ 2 / x ^ 2#?

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Answer 1 · 2016-07-01T12:36:50

#I=-(ln x)^2/x-(2ln x)/x-2/x+c#

The given function is

#(ln x) ^ 2 / x ^ 2#

We are to find out #I = int (ln x) ^ 2 / x ^ 2dx#

Let
#ln x =u =>x=e^u#

Differentiating w.r.t x we have

#(d(x))/(dx) =(d(e^u))/(du)*(du)/(dx) #

#=>1 =e^u*(du)/(dx) #

#=>dx=e^udu#

Now changing the variable the integral becomes

#I=intu^2/(e^(2u))*e^udu=intu^2*e^-udu#

#I= u^2inte^-udu-int((d(u^2))/(du)inte^-udu)du#

#= u^2inte^-udu-int2u(-e^-u)du#

#= u^2inte^-udu+int2ue^-udu#

#= u^2inte^-udu+2intue^-udu#

#= u^2inte^-udu+2uinte^-udu-2int((du)/(du)inte^-udu)du#

#= u^2inte^-udu+2uinte^-udu-2int(-e^-u)du#

#= u^2inte^-udu+2uinte^-udu+2int(e^-u)du#

#=-u^2e^-u-2ue^-u-2e^-u+c#

Inserting # u = lnx and e^-u =1/x#

#I=-(ln x)^2/x-(2ln x)/x-2/x+c#