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

$A=\frac{1}{3}$ $B=\frac{2}{3}$ $A=-\frac{1}{3}$ $B=-\frac{2}{3}$

$A=\frac{1}{3}$

$\int\frac{1}{(x^2+1)(x^2+4)}dx=\frac{1}{3}\int(\frac{1}{x^2+1}-\frac{1}{x^2+4})dx=\frac{1}{3}\tan^{-1}x-\frac{1}{6}\tan^{-1}\frac{x}{2}+C$ Therefore, $A=\frac{1}{3}$ and $B=-\frac{1}{6}$