# What is the antiderivative of (lnx)^2 / x^3?

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

#### Explanation:

Let
$y = {\left(\ln x\right)}^{2} / \left({x}^{3}\right)$

$\int y \mathrm{dx} = \int {\left(\ln x\right)}^{2} / \left({x}^{3}\right) \mathrm{dx}$
Let
$t = \ln x$

${e}^{t} = x$

${e}^{2} t = {x}^{2}$

${t}^{2} = {\left(\ln x\right)}^{2}$

$\frac{\mathrm{dt}}{\mathrm{dx}} = \frac{1}{x}$

$\mathrm{dt} = \frac{1}{x} \mathrm{dx}$

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

$= \int {t}^{2} / {e}^{2 t} \mathrm{dt}$

$\int y \mathrm{dx} = \int {t}^{2} {e}^{- 2 t} \mathrm{dt}$

integrating by parts

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

$u = {t}^{2}$

$\mathrm{du} = 2 t \mathrm{dt}$

$\mathrm{dv} = {e}^{-} 2 t \mathrm{dt}$

$v = - \frac{1}{2} {e}^{- 2 t}$

$\int {t}^{2} {e}^{- 2 t} \mathrm{dt} = {t}^{2} \left(- \frac{1}{2} {e}^{- 2 t}\right) - \int \left(- \frac{1}{2} {e}^{- 2 t}\right) \left(2 t \mathrm{dt}\right)$

$= - {t}^{2} / 2 {e}^{- 2 t} + \int t {e}^{- 2 t} \mathrm{dt}$

$= - {t}^{2} / 2 {e}^{- 2 t} + {I}_{1}$
where

${I}_{1} = \int t {e}^{- 2 t} \mathrm{dt}$

integrating by parts

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

$u = t$

$\mathrm{du} = \mathrm{dt}$

$\mathrm{dv} = {e}^{- 2 t} \mathrm{dt}$

$v = - \frac{1}{2} {e}^{- 2 t}$

$\int t {e}^{- 2 t} \mathrm{dt} = t \left(- \frac{1}{2} {e}^{- 2 t}\right) - \int \left(- \frac{1}{2} {e}^{- 2 t}\right) \mathrm{dt}$

$= - \frac{1}{2} t {e}^{- 2 t} + \frac{1}{2} \left(- \frac{1}{2}\right) {e}^{- 2 t}$

$\int t {e}^{- 2 t} \mathrm{dt} = - \frac{1}{2} t {e}^{- 2 t} - \frac{1}{4} {e}^{- 2 t}$

${I}_{1} = - \frac{1}{2} t {e}^{- 2 t} - \frac{1}{4} {e}^{- 2 t}$

$\int {t}^{2} {e}^{- 2 t} \mathrm{dt} = - {t}^{2} / 2 {e}^{- 2 t} + {I}_{1}$

$\int {t}^{2} {e}^{- 2 t} \mathrm{dt} = - {t}^{2} / 2 {e}^{- 2 t} + \left(- \frac{1}{2} t {e}^{- 2 t} - \frac{1}{4} {e}^{- 2 t}\right)$

$\int {t}^{2} {e}^{- 2 t} \mathrm{dt} = - {t}^{2} / 2 {e}^{- 2 t} - \frac{1}{2} t {e}^{- 2 t} - \frac{1}{4} {e}^{- 2 t}$

$\int {t}^{2} {e}^{- 2 t} \mathrm{dt} = - \frac{1}{4} \left(2 {t}^{2} + 2 t + 1\right) {e}^{-} 2 t$

Replacing

$t = \ln x$

${e}^{- 2 t} = \frac{1}{x} ^ 2$

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

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

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