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

Oct 19, 2016

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

#### Explanation:

You should learn the IBP formula:
$\int u \frac{\mathrm{dv}}{\mathrm{dx}} \mathrm{dx} = u v - \int v \frac{\mathrm{du}}{\mathrm{dx}} \mathrm{dx}$

So essentially we are looking for one function that simplifies when it is differentiated, and one that simplifies when integrated (or at least is integrable).

Hopefully you can spot that $\ln x$ is not easy to integrate (you need to using IBP again), but is simpler when differentiated.

Let $\left\{\begin{matrix}u = \ln x & \implies & \frac{\mathrm{du}}{\mathrm{dx}} = \frac{1}{x} \\ \frac{\mathrm{dv}}{\mathrm{dx}} = \frac{1}{x} ^ 3 = {x}^{-} 3 & \implies & v = {x}^{-} \frac{2}{-} 2 = - \frac{1}{2 {x}^{2}}\end{matrix}\right.$

So IBP gives;

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