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

Jul 1, 2016

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

#### Explanation:

The given function is

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

We are to find out $I = \int {\left(\ln x\right)}^{2} / {x}^{2} \mathrm{dx}$

Let
$\ln x = u \implies x = {e}^{u}$

Differentiating w.r.t x we have

$\frac{d \left(x\right)}{\mathrm{dx}} = \frac{d \left({e}^{u}\right)}{\mathrm{du}} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\implies 1 = {e}^{u} \cdot \frac{\mathrm{du}}{\mathrm{dx}}$

$\implies \mathrm{dx} = {e}^{u} \mathrm{du}$

Now changing the variable the integral becomes

$I = \int {u}^{2} / \left({e}^{2 u}\right) \cdot {e}^{u} \mathrm{du} = \int {u}^{2} \cdot {e}^{-} u \mathrm{du}$

using integration by parts

$I = {u}^{2} \int {e}^{-} u \mathrm{du} - \int \left(\frac{d \left({u}^{2}\right)}{\mathrm{du}} \int {e}^{-} u \mathrm{du}\right) \mathrm{du}$

$= {u}^{2} \int {e}^{-} u \mathrm{du} - \int 2 u \left(- {e}^{-} u\right) \mathrm{du}$

$= {u}^{2} \int {e}^{-} u \mathrm{du} + \int 2 u {e}^{-} u \mathrm{du}$

$= {u}^{2} \int {e}^{-} u \mathrm{du} + 2 \int u {e}^{-} u \mathrm{du}$

$= {u}^{2} \int {e}^{-} u \mathrm{du} + 2 u \int {e}^{-} u \mathrm{du} - 2 \int \left(\frac{\mathrm{du}}{\mathrm{du}} \int {e}^{-} u \mathrm{du}\right) \mathrm{du}$

$= {u}^{2} \int {e}^{-} u \mathrm{du} + 2 u \int {e}^{-} u \mathrm{du} - 2 \int \left(- {e}^{-} u\right) \mathrm{du}$

$= {u}^{2} \int {e}^{-} u \mathrm{du} + 2 u \int {e}^{-} u \mathrm{du} + 2 \int \left({e}^{-} u\right) \mathrm{du}$

$= - {u}^{2} {e}^{-} u - 2 u {e}^{-} u - 2 {e}^{-} u + c$

Inserting $u = \ln x \mathmr{and} {e}^{-} u = \frac{1}{x}$

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