If $\int \frac{x^2+2}{(x^2+1)(x^2+4)}dx=A{\tan }^{-1}x+B{\tan }^{-1}\frac{x}{2}+c$ ,then $(A,B)=$

$\begin{array}{l}I=\int \frac{x^2+2}{(x^2+1)(x^2+4)}dx\\ x^2+2=a(x^2+1)+b(x^2+4)\\ \Rightarrow a+b=1,a+4b=2\Rightarrow b=\frac{1}{3}\text{\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}and\hspace{0.17em}\hspace{0.17em}\hspace{0.17em}}a=\frac{2}{3}\end{array}$

$I=\int \frac{\frac{2}{3}(x^2+1)+\frac{1}{3}(x^2+4)}{(x^2+1)(x^2+4)}dx$

$=\frac{2}{3}\int \frac{1}{x^2+4}.dx+\frac{1}{3}\int \frac{1}{x^2+1}.dx$

$=\frac{2}{3}\times \frac{1}{2}{\tan }^{-1}\left(\frac{x}{2}\right)+\frac{1}{3}{\tan }^{-1}x+c$

$\therefore (A,B)=\left(\frac{{\displaystyle 1}}{{\displaystyle 3}},\frac{{\displaystyle 1}}{{\displaystyle 3}}\right)$