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

Jun 2, 2017

Begin with integration by parts:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

let $u = {\left(\ln \left(x\right)\right)}^{2}$ and $\mathrm{dv} = {x}^{-} 3 \mathrm{dx}$

then $\mathrm{du} = \frac{2 \ln \left(x\right)}{x} \mathrm{dx}$ and $v = - \frac{1}{2} {x}^{-} 2$

$\int {\left(\ln \left(x\right)\right)}^{2} / {x}^{3} \mathrm{dx} = - \frac{1}{2} {\left(\ln \frac{x}{x}\right)}^{2} - \int \left(- \frac{1}{2 {x}^{2}}\right) \left(\frac{2 \ln \left(x\right)}{x}\right) \mathrm{dx}$

$\int {\left(\ln \left(x\right)\right)}^{2} / {x}^{3} \mathrm{dx} = - \frac{1}{2} {\left(\ln \frac{x}{x}\right)}^{2} + \int \frac{\ln \left(x\right)}{x} ^ 3 \mathrm{dx}$

Integrate by parts:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

let $u = \ln \left(x\right)$ and $\mathrm{dv} = {x}^{-} 3 \mathrm{dx}$

then $\mathrm{du} = \frac{1}{x} \mathrm{dx}$ and $v = - \frac{1}{2} {x}^{-} 2$

$\int {\left(\ln \left(x\right)\right)}^{2} / {x}^{3} \mathrm{dx} = - \frac{1}{2} {\left(\ln \frac{x}{x}\right)}^{2} - \ln \frac{x}{2 {x}^{2}} - \int \left(- \frac{1}{2} {x}^{-} 2\right) \left(\frac{1}{x}\right) \mathrm{dx}$

$\int {\left(\ln \left(x\right)\right)}^{2} / {x}^{3} \mathrm{dx} = - \frac{1}{2} {\left(\ln \frac{x}{x}\right)}^{2} - \ln \frac{x}{2 {x}^{2}} + \frac{1}{2} \int \frac{1}{x} ^ 3 \mathrm{dx}$

We have already done the last integral:

$\int {\left(\ln \left(x\right)\right)}^{2} / {x}^{3} \mathrm{dx} = - \frac{1}{2} {\left(\ln \frac{x}{x}\right)}^{2} - \ln \frac{x}{2 {x}^{2}} - \frac{1}{4 {x}^{2}} + C$

Simplify over a common denominator:

$\int {\left(\ln \left(x\right)\right)}^{2} / {x}^{3} \mathrm{dx} = - \frac{2 {\left(\ln \left(x\right)\right)}^{2} + 2 \ln \left(x\right) + 1}{4 {x}^{2}} + C$