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

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

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

Put $x=-3\Rightarrow {(-3)}^2+5(-3)+1=A\left(0\right)+B\left(0\right)+C$

$9-15+1=C$

$C=-5$

Put $x=-2\Rightarrow 4+5(-2)+1=B(-2+3)+C$

$4-10+1=B-5$

$-5=B-5$

$\therefore B=0$