Answer
$$\frac{98}{3}$$
Work Step by Step
Given
$$\int_{0}^{4} x \sqrt{x^{2}+9} d x $$
Let
$$ u= x^2+9 \ \ \ \Rightarrow\ \ \ du = 2xdx$$
At $ x=0\to u= 9$ and $ x= 4 \to u= 25$
\begin{aligned} \int_{0}^{4} x \sqrt{x^{2}+9} d x &=\frac{1}{2} \int_{9}^{25} \sqrt{u} d u=\frac{1}{2} \int_{9}^{25} u^{1 / 2} d u \\ &=\left.\frac{1}{2}\left(\frac{2}{3} u^{3 / 2}\right)\right|_{9} ^{25} \\ &=\frac{1}{3}\left(25^{3 / 2}-9^{3 / 2}\right) \\ &=\frac{1}{3}(125-27)=\frac{98}{3}\end{aligned}