# How do you integrate int sqrtx ln 2x dx  using integration by parts?

Mar 7, 2018

$I = \frac{2}{9} \cdot {x}^{\frac{3}{2}} \left[3 \ln \left(2 x\right) - 2\right] + C$

#### Explanation:

$\textcolor{red}{I = \int u \cdot v \mathrm{dx} = u \cdot \int v \mathrm{dx} - \int \left({u}^{'} \cdot \int v \mathrm{dx}\right) \mathrm{dx}}$
$I = \int {x}^{\frac{1}{2}} \cdot \ln \left(2 x\right) \mathrm{dx} ,$
$T a k e , u = \ln \left(2 x\right) \mathmr{and} v = {x}^{\frac{1}{2}}$
$\Rightarrow {u}^{'} = \frac{1}{2 x} \cdot 2 = \frac{1}{x} \mathmr{and} \int v \mathrm{dx} = \frac{{x}^{\frac{3}{2}}}{\frac{3}{2}} = \frac{2}{3} \cdot {x}^{\frac{3}{2}} + c$
$I = \ln \left(2 x\right) \cdot \frac{2}{3} \cdot {x}^{\frac{3}{2}} - \int \frac{1}{x} \cdot \frac{2}{3} \cdot {x}^{\frac{3}{2}} \mathrm{dx}$
$I = \frac{2}{3} \cdot {x}^{\frac{3}{2}} \ln \left(2 x\right) - \frac{2}{3} \int {x}^{\frac{1}{2}} \mathrm{dx}$
$I = \frac{2}{3} \cdot {x}^{\frac{3}{2}} \ln \left(2 x\right) - \frac{2}{3} \cdot \frac{{x}^{\frac{3}{2}}}{\frac{3}{2}} + C$
$I = \frac{2}{3} \cdot {x}^{\frac{3}{2}} \ln \left(2 x\right) - \frac{4}{9} \cdot {x}^{\frac{3}{2}} + C$
$I = \frac{2}{9} \cdot {x}^{\frac{3}{2}} \left[3 \ln \left(2 x\right) - 2\right] + C$