# How do you find the integral (lnx)^2 / x^3?

Apr 29, 2018

$I = - \frac{1}{4 {x}^{2}} \left[2 {\left(\ln x\right)}^{2} + \ln x + 1\right] + c$

#### Explanation:

Here,

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

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

$\text{Using "color(blue)"Integration by Parts}$.

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

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

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

Again, $\text{using "color(blue)"Integration by Parts}$.

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

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

$= - {\left(\ln x\right)}^{2} / \left(2 {x}^{2}\right) - \frac{\ln x}{2 {x}^{2}} + \frac{1}{2} \left({x}^{- 2} / \left(- 2\right)\right) + c$

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

$I = - \frac{1}{4 {x}^{2}} \left[2 {\left(\ln x\right)}^{2} + \ln x + 1\right] + c$