If $\int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x=a x+b \tan ^{-1}\left(\frac{x}{\sqrt{3}}\right)+c \tan ^{-1}\left(\frac{x}{2}\right)+d$ (Where d is a constant of integration) then $\frac{a+c}{b}$ is equal to

$-\sqrt{3}$

$I = \int \frac{\left(x^2+1\right)\left(x^2+2\right)}{\left(x^2+3\right)\left(x^2+4\right)} d x$

$I =\int 1+\frac{2}{x^2+3}-\frac{6}{x^2+4} d x$

$=x+\frac{2}{\sqrt{3}} \tan ^{-1}\frac{x}{\sqrt{3}}-\frac{6}{2} \tan ^{-1} \frac{x}{2}+C$

so by partial function

Let Z = x²

$\frac{(z+1)(z+2)}{(z+3)(z+4)}=1+\frac{A}{z+3}+\frac{B}{z+4}$

so  $(z+1)(z+2)=(z+3)(z+4)+A(z+4)+B(z+3)$

at  z = -3

$A=-2 \times(-1) \Rightarrow A=+2$

at  z = -4

$-B=-3 \times(-2) \Rightarrow B=-6$

On comparison

a = 1    b = $\frac{2}{\sqrt{3}}$        c = -3

so   $\frac{a+c}{b} = \frac{-2}{\frac{2}{\sqrt{3}}} = -\sqrt{3}$

Option: A