# How do you integrate  [ln sqrt x] / x?

Apr 12, 2018

The integral equals $\frac{1}{4} {\ln}^{2} x + C$

#### Explanation:

We can rewrite using logarithm laws.

$I = \int \ln \frac{{x}^{\frac{1}{2}}}{x} \mathrm{dx}$

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

We now let $u = \ln x$. Then $\mathrm{du} = \frac{1}{x} \mathrm{dx}$ and then $\mathrm{dx} = x \mathrm{du}$.

$I = \int \frac{u}{2 x} \cdot x \mathrm{du}$

$I = \int \frac{1}{2} u \mathrm{du}$

$I = \frac{1}{2} \left(\frac{1}{2} {u}^{2}\right) + C$

$I = \frac{1}{4} {\ln}^{2} x + C$

Hopefully this helps!