If $\int \frac{x^2+4}{x^4+16}dx=\frac{1}{k}{\tan }^{-1}\left(\frac{{\displaystyle x^2-4}}{{\displaystyle kx}}\right)+C$ then k =

$\int \frac{x^2+4}{x^4+16}dx$

$Putx-\frac{4}{x}=t\Rightarrow \left(1+\frac{{\displaystyle 4}}{{\displaystyle x^2}}\right)dx=dt$

$=\int \left(\frac{1+\frac{4}{x^2}}{x^2+\frac{16}{x^2}}\right)dx$

$=\int \frac{dt}{t^2+8}$

$=\frac{1}{\sqrt{8}}{\tan }^{-1}\left(\frac{{\displaystyle t}}{{\displaystyle \sqrt{8}}}\right)+c$

$t^2={\left(x-\frac{{\displaystyle 4}}{{\displaystyle x}}\right)}^2\Rightarrow x^2+\frac{16}{x^2}-8\Rightarrow t^2+8=x^2+\frac{16}{x^2}$

$=\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{{\displaystyle x-\frac{{\displaystyle 4}}{{\displaystyle x}}}}{{\displaystyle \sqrt{8}}}\right)+c$

$\frac{1}{2\sqrt{2}}{\tan }^{-1}\left(\frac{{\displaystyle x^2-4}}{{\displaystyle 2\sqrt{2}x}}\right)+c$

$\therefore k=2\sqrt{2}$