How do you integrate $(ln x)^2/ x^3 dx$ ?

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Answer 1 · 2015-04-20T11:03:17

let's start by $u = ln(x)$
$du = 1/x$

$int ln^2(x)/x^3dx =int 1/x^2*1/x*ln^2(x) =int 1/x^2*u^2du$

$u = ln(x)$
$e^u=x$
$e^(2u) = x^2$
$e^(-2u)= 1/x^2$

So now we have :

$=>inte^(-2u)*u^2du$

By part :

$dv= e^(-2u)$
$v =-1/2e^(-2u)$

$w = u^2$
$dw = 2u$

$=>-1/2[u^2*e^(-2u)]+intu*e^(-2u)du$

By part again :

$dv = e^(-2u)$
$v = -1/2e^(-2u)$
$w = u$
$dw = 1$

$=>-1/2[u^2*e^(-2u)]-1/2[e^(-2u)*u]+1/2inte^(-2u)du$

$=>-1/2[u^2*e^(-2u)]-1/2[e^(-2u)*u]-1/4[e^(-2u)]$

Substitute back for $u = ln(x)$

$=>-1/2[ln^2(x)*1/x^2]-1/2[ln(x)*1/x^2]-1/4[1/x^2]+C$

$=>-1/(4x^2)(2ln^2(x)+2ln(x)+1)+C$

How do you integrate (ln x)^2/ x^3 dx(ln x)^2/ x^3 dx?