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

Jul 16, 2016

$I = \frac{1}{2} {\ln}^{2} x + C$

#### Explanation:

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

this can be done by inspection as its $\int \setminus f \left(x\right) f ' \left(x\right) \setminus \mathrm{dx} = \frac{1}{2} {f}^{2} \left(x\right) + C$

or we can use another method eg IBP......$\int u v ' = u v - \int u ' v$

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

$= \int \frac{d}{\mathrm{dx}} \left(\ln x\right) \setminus \ln x \setminus \mathrm{dx}$

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

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

$\implies I = {\ln}^{2} x - I + C$

$I = \frac{1}{2} {\ln}^{2} x + C$