What is #int (lnx)^2 / x^3dx#?

1 Answer
Jun 1, 2018

#I=-1/(4x^2)[2(lnx)^2+2lnx+1]+c#

Explanation:

Here,

#I=int(lnx)^2/x^3 dx=int(lnx)^2/x^2*1/xdx#

Let,

#lnx=u=>1/xdx=du#

#and x=e^u#

So,

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

#=int u^2*e^(-2u)du#

#"Using "color(blue)"Integration by Parts :"#

#I=u^2*inte^(-2u)du-int((2u)*inte^(-2u)du)du#

#=u^2(e^(-2u)/(-2))-int(2u)(e^(-2u)/(-2))du#

#=-u^2/(2e^(2u))+int(u*e^(-2u))du#

Again #"using "color(blue)"Integration by Parts :"#

#I=-u^2/(2e^(2u))+[u*(e^(-2u)/(-2))-int1*(e^(-2u)/(-2))du]#

#=-u^2/(2e^(2u))-u/(2e^(2u))+1/2inte^(-2u)du#

#=-u^2/(2e^(2u))-u/(2e^(2u))+1/2*(e^(-2u)/(-2))+c#

#=-u^2/(2e^(2u))-u/(2e^(2u))-1/(4e^(2u))+c#

#=-1/(4e^(2u))[2u^2+2u+1]+c#

Subst. back , #u=lnx ande^u=x#, we get

#I=-1/(4x^2)[2(lnx)^2+2lnx+1]+c#

NOTE :

#inte^(-2u)du# , we put , #-2u=t=>u=-t/2=>du=-1/2dt#

So,

#inte^(-2u)du=inte^t(-1/2)dt=-1/2inte^tdt=-1/2e^t=e^(-2u)/(-2)#