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, $"using "color(blue)"Integration by Parts"$ .

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

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

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

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

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

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