How do you find the integral #(lnx)^2 / x^3#?

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Answer 1 · 2018-04-29T12:18:25

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

Here,

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

#=int(lnx)^2*x^(-3)dx#

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

#I=(lnx)^2intx^(-3)dx-int(d/(dx)((lnx)^2)intx^(-3)dx)dx#

#=(lnx)^2(x^(-2)/(-2))-int2(lnx)*1/x(x^(-2)/(-2))dx#

#=-(lnx)^2/(2x^2)+int(lnx)(x^(-3))dx#

Again, #"using "color(blue)"Integration by Parts"#.

#I=-(lnx)^2/(2x^2)+[lnx(x^(-2)/(-2))-int1/x(x^(-2)/(-2))dx]#

#=-(lnx)^2/(2x^2)-(lnx)/(2x^2)+1/2intx^(-3)dx#

#=-(lnx)^2/(2x^2)-(lnx)/(2x^2)+1/2(x^(-2)/(-2))+c#

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

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