# How do you integrate (ln x)^2/ x^3 dx?

Apr 20, 2015

let's start by $u = \ln \left(x\right)$
$\mathrm{du} = \frac{1}{x}$

$\int {\ln}^{2} \frac{x}{x} ^ 3 \mathrm{dx} = \int \frac{1}{x} ^ 2 \cdot \frac{1}{x} \cdot {\ln}^{2} \left(x\right) = \int \frac{1}{x} ^ 2 \cdot {u}^{2} \mathrm{du}$

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

So now we have :

$\implies \int {e}^{- 2 u} \cdot {u}^{2} \mathrm{du}$

By part :

$\mathrm{dv} = {e}^{- 2 u}$
$v = - \frac{1}{2} {e}^{- 2 u}$

$w = {u}^{2}$
$\mathrm{dw} = 2 u$

$\implies - \frac{1}{2} \left[{u}^{2} \cdot {e}^{- 2 u}\right] + \int u \cdot {e}^{- 2 u} \mathrm{du}$

By part again :

$\mathrm{dv} = {e}^{- 2 u}$
$v = - \frac{1}{2} {e}^{- 2 u}$
$w = u$
$\mathrm{dw} = 1$

$\implies - \frac{1}{2} \left[{u}^{2} \cdot {e}^{- 2 u}\right] - \frac{1}{2} \left[{e}^{- 2 u} \cdot u\right] + \frac{1}{2} \int {e}^{- 2 u} \mathrm{du}$

$\implies - \frac{1}{2} \left[{u}^{2} \cdot {e}^{- 2 u}\right] - \frac{1}{2} \left[{e}^{- 2 u} \cdot u\right] - \frac{1}{4} \left[{e}^{- 2 u}\right]$

Substitute back for $u = \ln \left(x\right)$

$\implies - \frac{1}{2} \left[{\ln}^{2} \left(x\right) \cdot \frac{1}{x} ^ 2\right] - \frac{1}{2} \left[\ln \left(x\right) \cdot \frac{1}{x} ^ 2\right] - \frac{1}{4} \left[\frac{1}{x} ^ 2\right] + C$

$\implies - \frac{1}{4 {x}^{2}} \left(2 {\ln}^{2} \left(x\right) + 2 \ln \left(x\right) + 1\right) + C$