Answer
$9\pi$
Work Step by Step
\[\begin{align}
& \int_{1}^{\sqrt{7}}{2x\sqrt{36-{{\left( {{x}^{2}}-1 \right)}^{2}}}}dx \\
& \text{Let }u={{x}^{2}}-1,\text{ }du=2xdx \\
& \text{The new limits of integration are:} \\
& x=\sqrt{7}\to u={{\left( \sqrt{7} \right)}^{2}}-1=6 \\
& x=1\to u={{\left( 1 \right)}^{2}}-1=0 \\
& \text{Substituting} \\
& \int_{1}^{\sqrt{7}}{2x\sqrt{36-{{\left( {{x}^{2}}-1 \right)}^{2}}}}dx=\int_{0}^{6}{\sqrt{36-{{u}^{2}}}}du \\
& \text{By the equation of a semicirle} \\
& \int_{0}^{6}{\sqrt{36-{{u}^{2}}}}du=\frac{1}{2}\left( \frac{1}{2}\pi {{\left( 6 \right)}^{2}} \right) \\
& =9\pi \\
\end{align}\]
