Answer
$\dfrac{2}{3} x^{3/2} \ln x - \dfrac{4}{9} x^{3/2}+C $
Work Step by Step
Use integration by parts as follows:
$u \dfrac{dv}{dx} \ dx =uv -\int [\dfrac{du}{dx} v ] dx $
We have:
$u =\ln x \implies du=x^{-1} dx$ and $d v=x^{1/2} dx \implies v=\dfrac{2}{3} x^{3/2}$
Now,
$u \dfrac{dv}{dx} \ dx =uv -\int [\dfrac{du}{dx} v ] dx $
$I= \int \sqrt x \ln x \ dx \\=\dfrac{2}{3} x^{3/2} \ln x -\int \dfrac{2}{3} x^{1/2} \ dx \\= \dfrac{2}{3} x^{3/2} \ln x - \dfrac{4}{9} x^{3/2}+C $
