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

Answer 1

Samantha Adkison

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

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)#

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Answer 2

Emma Aldridge

The integral of (ln(x))^2 / x^3dx is given by:

∫ (ln(x))^2 / x^3 dx = - (ln(x))^2 / (2x^2) + (ln(x))/(2x^2) + C

Where C is the constant of integration.

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Answer from HIX Tutor

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