2x+1(x+1)(x−3)(x+4)=A(x+1) +B(x−3) +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
$C=-1/3$ $1/12+1/4-1/3=0$
