Answer
$$\frac{9}{68}$$
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=1\to u= 8$ and $ x= 2 \to u= 17$
\begin{aligned} \int_{1}^{2} \frac{4 x+12}{\left(x^{2}+6 x+1\right)^{2}} d x &=2 \int_{8}^{17} \frac{1}{u^{2}} d u\\
&=2 \int_{8}^{17} u^{-2} d u \\ &=-\left.2\left(\frac{1}{u}\right)\right|_{8} ^{17} \\ &=-2\left(\frac{1}{17}-\frac{1}{8}\right)=\frac{9}{68}\end{aligned}
