# How do you integrate lnx/x^.5?

May 26, 2016

$2 \sqrt{x} \left(\ln x - 2\right) + C$

#### Explanation:

This can be written as

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

Integration by parts takes the form:

$\int u \mathrm{dv} = u v - \int v \mathrm{du}$

Here, let $u = \ln x$ and $\mathrm{dv} = \frac{1}{\sqrt{x}} \mathrm{dx}$.

These imply that:

$\left\{\begin{matrix}u = \ln x \text{ "=>" "du=1/xdx \\ dv=x^(-1/2)dx" "=>" } v = 2 {x}^{\frac{1}{2}} = 2 \sqrt{x}\end{matrix}\right.$

Thus,

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

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

$= 2 \sqrt{x} \ln x - 4 \sqrt{x} + C$

$= 2 \sqrt{x} \left(\ln x - 2\right) + C$