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

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Answer 1 · 2017-06-02T17:46:48

$intudv = uv - intvdu$

let $u = (ln(x))^2$ and $dv = x^-3dx$

then $du = (2ln(x))/xdx$ and $v = -1/2x^-2$

$int(ln(x))^2/x^3dx = -1/2(ln(x)/x)^2 -int(-1/(2x^2))((2ln(x))/x)dx$

$int(ln(x))^2/x^3dx = -1/2(ln(x)/x)^2 +int(ln(x))/x^3dx$

Integrate by parts:

$intudv = uv - intvdu$

let $u = ln(x)$ and $dv = x^-3dx$

then $du = 1/xdx$ and $v = -1/2x^-2$

$int(ln(x))^2/x^3dx = -1/2(ln(x)/x)^2 - ln(x)/(2x^2) - int (-1/2x^-2)(1/x)dx$

$int(ln(x))^2/x^3dx = -1/2(ln(x)/x)^2 - ln(x)/(2x^2) +1/2int 1/x^3dx$

We have already done the last integral:

$int(ln(x))^2/x^3dx = -1/2(ln(x)/x)^2 - ln(x)/(2x^2) -1/(4x^2)+ C$

Simplify over a common denominator:

$int(ln(x))^2/x^3dx = -(2(ln(x))^2 + 2ln(x) +1)/(4x^2)+ C$

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