# Find  I = int \ lnx/x^2 \ dx ?

Oct 11, 2017

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

#### Explanation:

We seek:

$I = \int \setminus \ln \frac{x}{x} ^ 2 \setminus \mathrm{dx}$

We can apply integration by by parts:

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

Then plugging into the IBP formula:

$\int \setminus \left(u\right) \left(\frac{\mathrm{dv}}{\mathrm{dx}}\right) \setminus \mathrm{dx} = \left(u\right) \left(v\right) - \int \setminus \left(v\right) \left(\frac{\mathrm{du}}{\mathrm{dx}}\right) \setminus \mathrm{dx}$

gives us

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

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

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

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