# How do you integrate int lnx/x^2 by parts?

Jul 30, 2017

The answer is $= - \frac{\ln x}{x} - \frac{1}{x} + C$

#### Explanation:

The integration by parts is

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

Here,

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

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

Therefore,

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