[tex]E(x)=\frac{4}{x^2+8}-\frac{3}{x^2+5}\\
E(x):(2x^2+16)^{-1}=\\
\left(\frac{4}{x^2+8}-\frac{3}{x^2+5}\right):\frac{1}{2(x^2+8)}=\\
\frac{4(x^2+5)-3(x^2+8)}{(x^2+8)(x^2+5)}\cdot 2(x^2+8)=\\
\frac{4x^2+20-3x^2-24}{x^2+5}\cdot 2=\\
\frac{2(x^2-4)}{x^2+5}=\boxed{\boxed{\frac{2(x-2)(x+2)}{x^2+5}}}[/tex]