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#