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

\(A=\frac{1}{3}, B=-\frac{2}{3}\) \(A=\frac{1}{3}, B=-\frac{1}{6}\) \(A=\frac{1}{3}, B=\frac{2}{3}\) \(A=\frac{1}{6}, B=-\frac{2}{3}\)

\(A=\frac{1}{3}, B=-\frac{1}{6}\)

Assume \(x^2=t\) and apply partial fraction