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

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Answer 1 · 2016-12-09T15:45:44

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

We have

$I = int (lnx)^2/x^2dx$

If $z = lnx$ then $dz = dx/x$ and

$z = ln x$ so $e^z = x$

$I = int z^2e^(-z)dz$

Integrating by parts

$u = z^2$ so $du = 2z dz$
$dv = e^(-z) dz$ so $v = -e^(-z)$

$I = u*v - int v du$
$I = -z^2e^(-z) + 2intze^(-z)dz$

Integrating by parts once again

$u = z$ so $du = dz$
$dv = e^(-z)dz$ so $v = -e^(-z)$

$I = -z^2e^(-z) + 2(-ze^(-z) + inte^(-z)dz)$
$I = -z^2e^(-z) - 2ze^(-z) - 2e^(-z) + c$
$I = -e^(-z)(z^2 + 2z + 2) + c$

But we want to have an answer in terms of $x$ so if we remember that $z = ln(x)$ we'll have

$I = -e^(-ln(x))((lnx)^2 + 2lnx + 2) +c$
$I = c-((lnx)^2 + 2lnx + 2)/x$

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