# How do you integrate int sqrtxlnx by integration by parts method?

Jan 29, 2017

$\int \sqrt{x} \ln \left(x\right) \mathrm{dx} = \frac{2}{3} {x}^{\frac{3}{2}} \left(\ln \left(x\right) - \frac{2}{3}\right) + C$

#### Explanation:

Using integration by parts:

Let $u = \ln \left(x\right) \implies \mathrm{du} = \frac{1}{x} \mathrm{dx}$
and $\mathrm{dv} = {x}^{\frac{1}{2}} \mathrm{dx} \implies v = \frac{2}{3} {x}^{\frac{3}{2}}$

Using the integration by parts formula $\int u \mathrm{dv} = u v - \int v \mathrm{du}$

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

$= \frac{2}{3} {x}^{\frac{3}{2}} \ln \left(x\right) - \int \frac{2}{3} {x}^{\frac{3}{2}} \cdot \frac{1}{x} \mathrm{dx}$

$= \frac{2}{3} {x}^{\frac{3}{2}} \ln \left(x\right) - \frac{2}{3} \int {x}^{\frac{1}{2}} \mathrm{dx}$

$= \frac{2}{3} {x}^{\frac{3}{2}} \ln \left(x\right) - \frac{2}{3} \left(\frac{2}{3} {x}^{\frac{3}{2}}\right) + C$

$= \frac{2}{3} {x}^{\frac{3}{2}} \left(\ln \left(x\right) - \frac{2}{3}\right) + C$