How do you integrate $\frac{{(ln x)}^{2}}{x^{3}} d x$ $(ln x)^2/ x^3 dx$ ?

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How do you integrate $\frac{{(ln x)}^{2}}{x^{3}} d x$ $(ln x)^2/ x^3 dx$ ?

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Answer 1 · 2015-04-20T11:03:17

let's start by $u = ln (x)$ $u = ln(x)$
$d u = \frac{1}{x}$ $du = 1/x$

$\int \frac{{ln}^{2} (x)}{x^{3}} d x = \int \frac{1}{x^{2}} \cdot \frac{1}{x} \cdot {ln}^{2} (x) = \int \frac{1}{x^{2}} \cdot u^{2} d u$ $int ln^2(x)/x^3dx =int 1/x^2*1/x*ln^2(x) =int 1/x^2*u^2du$

$u = ln (x)$ $u = ln(x)$
$e^{u} = x$ $e^u=x$
$e^{2 u} = x^{2}$ $e^(2u) = x^2$
$e^{- 2 u} = \frac{1}{x^{2}}$ $e^(-2u)= 1/x^2$

So now we have :

$\Rightarrow \int e^{- 2 u} \cdot u^{2} d u$ $=>inte^(-2u)*u^2du$

By part :

$d v = e^{- 2 u}$ $dv= e^(-2u)$
$v = - \frac{1}{2} e^{- 2 u}$ $v =-1/2e^(-2u)$

$w = u^{2}$ $w = u^2$
$d w = 2 u$ $dw = 2u$

$\Rightarrow - \frac{1}{2} [u^{2} \cdot e^{- 2 u}] + \int u \cdot e^{- 2 u} d u$ $=>-1/2[u^2*e^(-2u)]+intu*e^(-2u)du$

By part again :

$d v = e^{- 2 u}$ $dv = e^(-2u)$
$v = - \frac{1}{2} e^{- 2 u}$ $v = -1/2e^(-2u)$
$w = u$ $w = u$
$d w = 1$ $dw = 1$

$\Rightarrow - \frac{1}{2} [u^{2} \cdot e^{- 2 u}] - \frac{1}{2} [e^{- 2 u} \cdot u] + \frac{1}{2} \int e^{- 2 u} d u$ $=>-1/2[u^2*e^(-2u)]-1/2[e^(-2u)*u]+1/2inte^(-2u)du$

$\Rightarrow - \frac{1}{2} [u^{2} \cdot e^{- 2 u}] - \frac{1}{2} [e^{- 2 u} \cdot u] - \frac{1}{4} [e^{- 2 u}]$ $=>-1/2[u^2*e^(-2u)]-1/2[e^(-2u)*u]-1/4[e^(-2u)]$

Substitute back for $u = ln (x)$ $u = ln(x)$

$\Rightarrow - \frac{1}{2} [{ln}^{2} (x) \cdot \frac{1}{x^{2}}] - \frac{1}{2} [ln (x) \cdot \frac{1}{x^{2}}] - \frac{1}{4} [\frac{1}{x^{2}}] + C$ $=>-1/2[ln^2(x)*1/x^2]-1/2[ln(x)*1/x^2]-1/4[1/x^2]+C$

$\Rightarrow - \frac{1}{4 x^{2}} (2 {ln}^{2} (x) + 2 ln (x) + 1) + C$ $=>-1/(4x^2)(2ln^2(x)+2ln(x)+1)+C$

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