Suppose that, #I=int(x^2-4)/{(x+3)(x^2+1)}dx#.
Let, #(x^2-4)/{(x+3)(x^2+1)}=A/(x+3)+(Bx+C)/(x^2+1), (A,B,C in RR)...(ast)#.
We will use Heavyside's Method to determine #A,B,C#.
#:. A=[(x^2-4)/(x^2+1)]_(x=-3)=(9-4)/(9+1)=1/2#.
Using #A# in #(ast)#, we have,
# (Bx+C)/(x^2+1)=(x^2-4)/{(x+3)(x^2+1)}-(1/2)/(x+3)#,
#={2(x^2-4)-(x^2+1)}/{2(x+3)(x^2+1)}, i.e., #
# (Bx+C)/(x^2+1)=(x^2-9)/{2(x+3)(x^2+1)}={1/2(x-3)}/(x^2+1)#.
#:. B=1/2, and, C=-3/2#.
Altogether, we have,
#(x^2-4)/{(x+3)(x^2+1)}=(1/2)/(x+3)+(1/2x-3/2)/(x^2+1)#.
#:. I=1/2int1/(x+3)dx+1/2intx/(x^2+1)dx-3/2int1/(x^2+1)dx#,
#=1/2ln|(x+3)|+1/4int(2x)/(x^2+1)dx-3/2arc tanx#,
#=1/2ln|(x+3)|+1/4int{d/dx(x^2+1)}/(x^2+1)dx-3/2arc tanx#,
#rArr I=1/2ln|(x+3)|+1/4ln(x^2+1)-3/2arc tanx+C#.
Enjoy Maths.!