# How do you integrate (ln(x)*(1/x))dx?

Jul 3, 2016

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

#### Explanation:

$\int \mathrm{dx} q \quad \ln \left(x\right) \cdot \frac{1}{x}$

you can do this by inspection as $\left(\ln x\right) ' = \frac{1}{x}$ so we can trial $\alpha {\left(\ln x\right)}^{2}$ as a solution.

So $\left(\alpha {\left(\ln x\right)}^{2} + C\right) ' = 2 \alpha \ln x \frac{1}{x} \implies 2 \alpha = 1 , \alpha = \frac{1}{2}$

if you don't fancy that you could use IBP : $\int u v ' = u v - \int u ' v$

$u = \ln x , u ' = \frac{1}{x}$
$v ' = \frac{1}{x} , v = \ln x$

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

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

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