$\int \dfrac{d x}{(x+2)\left(x^2+1\right)}=p \log |x+2|+q \log \left|x^2+1\right|+r \tan ^{-1} x+c$ then $p+q+r=$

Let $\dfrac{1}{(x+2)(x^2+1)} = \dfrac{A}{x+2} + \dfrac{Bx+C}{x^2+1}$.

Multiplying both sides by $(x+2)(x^2+1)$, we get $1 = A(x^2+1) + (Bx+C)(x+2)$.

Setting $x = -2$, we get $1 = A(4+1) + 0$, which implies $5A = 1$, so $A = \dfrac{1}{5}$.

Comparing coefficients of $x^2$, we get $A + B = 0$, so $B = -A = -\dfrac{1}{5}$.

Comparing constant terms, we get $A + 2C = 1$, so $\dfrac{1}{5} + 2C = 1$, which implies $2C = \dfrac{4}{5}$, so $C = \dfrac{2}{5}$.

The integral becomes $\int \left( \dfrac{1/5}{x+2} + \dfrac{-x/5 + 2/5}{x^2+1} \right) dx = \dfrac{1}{5} \int \dfrac{dx}{x+2} - \dfrac{1}{10} \int \dfrac{2x}{x^2+1} dx + \dfrac{2}{5} \int \dfrac{dx}{x^2+1}$.

This evaluates to $\dfrac{1}{5} \log |x+2| - \dfrac{1}{10} \log |x^2+1| + \dfrac{2}{5} \tan^{-1} x + c$.

Comparing with $p \log |x+2| + q \log |x^2+1| + r \tan^{-1} x + c$, we have $p = \dfrac{1}{5}$, $q = -\dfrac{1}{10}$, and $r = \dfrac{2}{5}$.

Therefore, $p + q + r = \dfrac{1}{5} - \dfrac{1}{10} + \dfrac{2}{5} = \dfrac{2 - 1 + 4}{10} = \dfrac{5}{10} = \dfrac{1}{2}$.

Answer: $\dfrac{1}{2}$