Answer
$= \frac{2}{3}\sqrt {x^{3}}(\ln x - \frac{2}{3}) + C$
Work Step by Step
$\int \sqrt x \ln x dx$
$u = \ln x$
$u' = \frac{1}{x}$
$\frac{du}{dx} = \frac{1}{x}$
$du = \frac{dx}{x}$
$dv = \sqrt x dx$
$v = \frac{x^{\frac{3}{2}}}{\frac{3}{2}}$
$v = \frac{2}{3}x^{\frac{3}{2}}$
$uv - \int vdu$
$= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \int (\frac{2}{3}x^{\frac{3}{2}})(\frac{dx}{x})$
$= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \frac{2}{3}\int (x^{\frac{3}{2}})(\frac{1}{x})dx$
$= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \frac{2}{3}\int (x^{\frac{3}{2}})(x^{-1})dx$
$= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \frac{2}{3}\int (x^{\frac{1}{2}})dx$
$= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \frac{2}{3}\frac{2}{3}x^{\frac{3}{2}}$
$= (\ln x)(\frac{2}{3}x^{\frac{3}{2}}) - \frac{4}{9}x^{\frac{3}{2}}$
$= \frac{2}{3}x^{\frac{3}{2}}(\ln x - \frac{2}{3}) + C$
$= \frac{2}{3}\sqrt {x^{3}}(\ln x - \frac{2}{3}) + C$
