How do you integrate $\frac{{(ln x)}^{2}}{x^{2}}$ $(ln x) ^ 2 / x ^ 2$ ?

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How do you integrate $\frac{{(ln x)}^{2}}{x^{2}}$ $(ln x) ^ 2 / x ^ 2$ ?

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Answer 1 · 2016-12-09T15:45:44

$I = c - \frac{{(ln x)}^{2} + 2 ln x + 2}{x}$ $I = c-((lnx)^2 + 2lnx + 2)/x$

Explanation:

We have

$I = \int \frac{{(ln x)}^{2}}{x^{2}} d x$ $I = int (lnx)^2/x^2dx$

If $z = ln x$ $z = lnx$ then $d z = \frac{d x}{x}$ $dz = dx/x$ and

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

$I = \int z^{2} e^{- z} d z$ $I = int z^2e^(-z)dz$

Integrating by parts

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

$I = u \cdot v - \int v d u$ $I = u*v - int v du$
$I = - z^{2} e^{- z} + 2 \int z e^{- z} d z$ $I = -z^2e^(-z) + 2intze^(-z)dz$

Integrating by parts once again

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

$I = - z^{2} e^{- z} + 2 (- z e^{- z} + \int e^{- z} d z)$ $I = -z^2e^(-z) + 2(-ze^(-z) + inte^(-z)dz)$
$I = - z^{2} e^{- z} - 2 z e^{- z} - 2 e^{- z} + c$ $I = -z^2e^(-z) - 2ze^(-z) - 2e^(-z) + c$
$I = - e^{- z} (z^{2} + 2 z + 2) + c$ $I = -e^(-z)(z^2 + 2z + 2) + c$

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

$I = - e^{- ln (x)} ({(ln x)}^{2} + 2 ln x + 2) + c$ $I = -e^(-ln(x))((lnx)^2 + 2lnx + 2) +c$
$I = c - \frac{{(ln x)}^{2} + 2 ln x + 2}{x}$ $I = c-((lnx)^2 + 2lnx + 2)/x$

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How do you integrate $\frac{{(ln x)}^{2}}{x^{2}}$ $(ln x) ^ 2 / x ^ 2$ ?

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