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

$-\frac{5}{2}\tan^{-1}x+\frac{6}{\sqrt{2}}\tan^{-1}(\frac{x}{\sqrt{2}})-\frac{7}{2\sqrt{3}}\tan^{-1}(\frac{x}{\sqrt{3}})+C$

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

let $x^2=y$

so $\frac{y-4}{(y+1)(y+2)(y+3)}=\frac{A}{y+1}+\frac{B}{y+2}+\frac{C}{y+3}$

so $t-4=A(y+2)(y+3)+B(y+1)(y+3)+C(y+1)(y+2)$

at $y=-1$,  $-5=2A⇒A=-\frac{5}{2}$

at $y=-2$,  $-3-3=-B⇒B=6$

at $y=-3$,  $-7=C(2)⇒C=-\frac{7}{2}$

$I=\int-\frac{5}{2(x^2+1)}+\frac{6}{(x^2+2)}-\frac{7}{2(x^2+3)}dx$

so $I=-\frac{5}{2}\tan^{-1}(x)+\frac{6}{\sqrt{2}}\tan^{-1}\frac{x}{\sqrt{2}}-\frac{7}{2\sqrt{3}}\tan^{-1}\frac{x}{\sqrt{3}}+C$