Given,

$x^3+4x+1=0$

$a,b \& c$ are the roots of the given equation therefore

$a+b+c=0 ...\left(i\right)$

$ab+bc+ca=4 ...\left(ii\right)$

$abc=-1 ...\left(iii\right)$

From the equation $\left(i\right),$ we have

$a+b=-c, b+c=-a \& c+a=-b$

Now,

$\frac{1}{a+b}+\frac{1}{b+c}+\frac{1}{c+a}$

$=\left(-\frac{1}{c}\right)+\left(-\frac{1}{a}\right)+\left(-\frac{1}{b}\right)$

$=-\frac{ab+bc+ca}{abc}$

$=4$.