\[\dfrac{2x+1}{(x+1)(x-3)(x+4)} = \dfrac{A}{(x+1)}\ + \dfrac{B}{(x-3)}\ + \dfrac{C}{(x+4)}\]
The above rational fraction is expressed in the form of partial fractions. Evaluate $A + B + C$.
We can use fraction decomposition,
$A(x-3)(x+4)+B(x+1)(x+4)+C(x+1)(x-3)=2x+1$
Setting $x$ as $-1$, we get
$A(-4)(3)=-1$
$A=1/12$
setting $x$ as $3$,
we get
B(4)(7)=7
$B=1/4$
Setting x=-4
C(-3)(-7)=-7
$C=-1/3$ $1/12+1/4-1/3=0$
