Answer
$4(\sqrt5-2)$
Work Step by Step
The area is indicated by the purple lines. We need to first calculate the intersection points.
$\frac{x}{\sqrt{1+x^2}}=\frac{x}{\sqrt{9-x^2}}$
$1+x^2=9-x^2$
$x=2;x=0;x=-2$
$\int_{-2}^0\Big(\frac{x}{\sqrt{9-x^2}}-\frac{x}{\sqrt{1+x^2}}\Big)dx+\int_{0}^2\Big(\frac{x}{\sqrt{1+x^2}}-\frac{x}{\sqrt{9-x^2}}\Big)dx$
$\int\frac{x}{\sqrt{9-x^2}}dx$
$u=9-x^2;\frac{du}{dx}=-2x;dx=\frac{du}{-2x}$
$\int\frac{x}{\sqrt{u}}\frac{du}{-2x}=\frac{-1}{2}\int\frac{1}{\sqrt{u}}du=-\sqrt{u}$
$\int\frac{x}{\sqrt{9-x^2}}dx=-\sqrt{9-x^2}$
Similarly, $\int\frac{x}{\sqrt{1+x^2}}=\sqrt{1+x^2}$
$\int_{-2}^0\Big(\frac{x}{\sqrt{9-x^2}}-\frac{x}{\sqrt{1+x^2}}\Big)dx+\int_{0}^2\Big(\frac{x}{\sqrt{1+x^2}}-\frac{x}{\sqrt{9-x^2}}\Big)dx=$
$[-\sqrt{9-x^2}-\sqrt{1+x^2}]_{-2}^{0}+[\sqrt{1+x^2}+\sqrt{9-x^2}]_0^2=4(\sqrt5-2)$
