# Question #bc6c7

Aug 27, 2016

Integrate by parts and exploit the fact the integral repeats itself. Answer is:

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

#### Explanation:

The integral is equal to:

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

Use $\sqrt{x} = {x}^{\frac{1}{2}}$ and $\ln {x}^{a} = a \ln x$

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

$= \frac{1}{2} \int \ln x \cdot \left(\ln x\right) ' \mathrm{dx}$

$I = \frac{1}{2} \left(\left(\ln x \cdot \ln x\right) - \int \left(\ln x\right) ' \cdot \ln x \mathrm{dx}\right) =$

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

$= \frac{1}{2} {\ln}^{2} x - \int \ln {x}^{\frac{1}{2}} / x \mathrm{dx} = \frac{1}{2} {\ln}^{2} x - \int \ln \frac{\sqrt{x}}{x} \mathrm{dx}$

Noticing the last integral is equal to the original:

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

Therefore:

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

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

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

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

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