# What is the antiderivative of ln(x)/x?

Apr 1, 2015

${\left(\ln \left(x\right)\right)}^{2} / 2 + C$ (Also written $\frac{1}{2} {\ln}^{2} \left(x\right) + c$).

Explanation:

$\ln \frac{x}{x} = \ln \left(x\right) \cdot \left(\frac{1}{x}\right)$.

The derivative of $\ln x$ is $\frac{1}{x}$, so

$\ln \left(x\right) \cdot \left(\frac{1}{x}\right)$ is of the form: $f \left(x\right) \cdot f ' \left(x\right)$, so the antiderivative is $\frac{1}{2} {\left(f \left(x\right)\right)}^{2} + C$.

(Alt notation: $\ln \frac{x}{x} = \ln \left(x\right) \cdot \left(\frac{1}{x}\right)$ is of the form $u \frac{\mathrm{du}}{\mathrm{dx}}$ whose antiderivative is ${u}^{2} / 2 + C$.)