If $\frac{x^2+5x+1}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}=\frac{A}{x+1}+\frac{B}{\left(x+1\right)\left(x+2\right)}+\frac{C}{\left(x+1\right)\left(x+2\right)\left(x+3\right)}$  then $B=$

$x^2+5x+1=A\left(x+2\right)\left(x+3\right)+B\left(x+3\right)+C$

Put $x=2$

$4-10+1=B+C\Rightarrow -5=B+C\to \left(1\right)$

Put $x=-3$

$9-15+1=C\Rightarrow C=-5\to \left(2\right)$

$B-5=-5=0$