## Calculus 8th Edition

$\int\sqrt{x}\ln{x}dx$
Let $u=\ln{x}$ and $dv=\sqrt{x}dx$ then $du=\frac{1}{x}dx$ and $v=\frac{2}{3}x^\frac{3}{2}$ then $\int\sqrt{x}\ln{x}dx=\sqrt{x}\ln{x}-\int\frac{2}{3}x^\frac{3}{2}\times\frac{1}{x}dx=\sqrt{x}\ln{x}-\int\frac{2}{3}x^\frac{1}{2}dx$ \begin{equation*}=\sqrt{x}\ln{x}-\frac{4}{9}x^\frac{3}{2}+C\end{equation*}