Definite integral via u-substitution

Let $u = x^2+9$. Then $du = 2x\,dx$, so $x\,dx = \tfrac{1}{2}du$. When $x=0$, $u=9$; when $x=4$, $u = 16+9=25$. Substituting, the integral becomes $\int_9^{25} \sqrt{u}\cdot \tfrac{1}{2}\,du = \tfrac{1}{2}\int_9^{25} u^{1/2}\,du.$

Integrating, $\tfrac{1}{2}\int_9^{25} u^{1/2}\,du = \tfrac{1}{2}\cdot \tfrac{2}{3}u^{3/2}\Big|_9^{25} = \tfrac{1}{3}\left(25^{3/2} - 9^{3/2}\right)$. Since $25^{3/2}=125$ and $9^{3/2}=27$, this equals $\tfrac{1}{3}(125-27) = \tfrac{98}{3}$.