# How do you integrate int (lnx)^2/x^3 using integration by parts?

Jul 30, 2017

The answer is $= - {\left(\ln x\right)}^{2} / \left(2 {x}^{2}\right) - \ln \frac{x}{2 {x}^{2}} - \frac{1}{4 {x}^{2}} + C$

#### Explanation:

The integration by oarts is

$\int u v ' \mathrm{dx} = u v - \int u ' v \mathrm{dx}$

Here,

$u = {\left(\ln x\right)}^{2}$, $\implies$, $u ' = 2 \ln x \cdot \frac{1}{x}$

$v ' = \frac{1}{x} ^ 3$, $\implies$, $v = - \frac{1}{2 {x}^{2}}$

Therefore,

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

We do once more the integration by parts

$u = \ln x$,$\implies$, $u ' = \frac{1}{x}$

$v ' = \frac{1}{x} ^ 3$, $\implies$, $v = - \frac{1}{2 {x}^{2}}$

So,

$\int \frac{\ln x}{x} ^ 3 = - \ln \frac{x}{2 {x}^{2}} + \int \frac{\mathrm{dx}}{2 {x}^{3}} = - \ln \frac{x}{2 {x}^{2}} - \frac{1}{4 {x}^{2}}$

Therefore,

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