If $\int \frac{4 x+1}{x^2+3 x+2} d x=a \log |x+1|+b \log |x+2|+C$, then

$a+b=4$

We have,

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

$\Rightarrow I=\int \frac{2(2 x+3)-5}{x^2+3 x+2} d x$         [Using $4 x+1=\lambda(2 x+3)+\mu$]

$\Rightarrow I=2 \int \frac{2 x+3}{x^2+3 x+2} d x-5 \int \frac{1}{x^2+3 x+2} d x$

$\Rightarrow I=2 \int \frac{1}{x^2+3 x+2} d\left(x^2+3 x+2\right)-5 \int \frac{1}{\left(x+\frac{3}{2}\right)^2-\left(\frac{1}{2}\right)^2} d x$

$\Rightarrow I=2 \log \left|x^2+3 x+2\right|-5 \log \left|\frac{x+1}{x+2}\right|+C$

$\Rightarrow I=2 \log |x+1|+2 \log |x+2|-5 \log |x+1|$

$\Rightarrow I=-3 \log |x+1|+7 \log |x+2|+C$

∴  $a=-3$ and $b=7 \Rightarrow a+b=4$