# How do I find the integral intln(x)/sqrt(x)dx ?

Jul 26, 2014

You can proceed by parts , that is, using the formula
$\setminus \int u \mathrm{dv} = u v - \setminus \int v \mathrm{du}$. See integration by parts

This means that you need to pick what you will interpret as $u$ (to be differentiated) and what you will interpret as $\mathrm{dv}$ (to be integrated).

Since you do not know (or, I should say, are usually not expected to know but rather to know how to obtain: by parts) an antiderivative of $\ln x$ but you know its derivative, you have no choice: you differentiate $\ln x$ and integrate $\frac{1}{\sqrt{x}} = {x}^{- \frac{1}{2}}$.

In other words,
$u = \ln x$ so that $\mathrm{du} = \frac{\mathrm{dx}}{x}$ and $\mathrm{dv} = {x}^{-} \left(\frac{1}{2}\right)$ so that $v = \setminus \int {x}^{- \frac{1}{2}} \mathrm{dx} = 2 {x}^{\frac{1}{2}}$. Applying the formula above
$\setminus \int \setminus \ln \frac{x}{\sqrt{x}} \mathrm{dx} = 2 \sqrt{x} \ln x - 2 \setminus \int {x}^{\frac{1}{2}} / x \mathrm{dx} = 2 \sqrt{x} \ln x - 2 \setminus \int {x}^{- \frac{1}{2}} \mathrm{dx}$ so that
$\setminus \int \setminus \ln \frac{x}{\sqrt{x}} \mathrm{dx} = 2 \sqrt{x} \ln x - 4 \sqrt{x} + C$