## University Calculus: Early Transcendentals (3rd Edition)

$$\int\sqrt{x}\ln xdx=\frac{2}{3}x^{3/2}\ln x-\frac{4}{9}x^{3/2}+C$$
$$A=\int\sqrt{x}\ln xdx$$ Take $u=\ln x$ and $dv=\sqrt xdx=x^{1/2}dx$ Then we have, $du=\frac{dx}{x}$ and $v=\frac{x^{3/2}}{\frac{3}{2}}=\frac{2x^{3/2}}{3}$ Applying the formula $\int udv=uv-\int vdu$, we have $$A=\frac{2}{3}x^{3/2}\ln x-\frac{2}{3}\int x^{3/2}\times\frac{dx}{x}$$ $$A=\frac{2}{3}x^{3/2}\ln x-\frac{2}{3}\int x^{1/2}dx$$ $$A=\frac{2}{3}x^{3/2}\ln x-\frac{2}{3}\times\frac{2x^{3/2}}{3}+C$$ $$A=\frac{2}{3}x^{3/2}\ln x-\frac{4}{9}x^{3/2}+C$$