Let's find such numbers $a, b, p$, and $q$ for which the following equality holds identically:

$$
\frac{6 x+1}{6 x^{2}+19 x+15}=\frac{a}{x-p}+\frac{b}{x-q}
$$
After simplifying the fractions, the right side is

$$
\frac{(a+b) x-(a q+b p)}{x^{2}-(p+q) x+p q}
$$

When the fraction is extended by 6, the right side becomes

$$
\frac{6(a+b) x-6(a q+b p)}{6 x^{2}-6(p+q) x+6 p q}
$$

Since the coefficient of $x^{2}$ in the denominator is 6 on both sides, the desired identity can only hold if the numerators and denominators of the two fractions are identical. That is, if

$$
6 x+1=6(a+b) x-6(a q+b p)
$$

and

$$
6 x^{2}+19 x+15=6 x^{2}-6(p+q)+6 p q
$$

From this, it follows that the coefficients of the appropriate powers of $x$ are equal, that is,

$$
a+b=1 \quad(1), \quad a q+b p=\frac{1}{6} \quad(2), \quad p+q=-\frac{19}{6} \quad(3), \quad p q=\frac{5}{2}
$$

From (3) and (4), it follows that $p$ and $q$ are the two roots of the following quadratic equation

$$
u^{2}+\frac{19}{6} u+\frac{5}{2}=0
$$

From this,

$$
p=-\frac{3}{4}, \quad q=-\frac{5}{3}
$$

(Swapping $p$ and $q$ only results in the values of $a$ and $b$ being swapped, so we do not get a substantially different solution.)

Substituting the values of $p$ and $q$ into equation (2),

$$
-\frac{5 a}{3}-\frac{3 b}{2}=-\frac{1}{6} \quad \text { that is } \quad 10 a+9 b=1
$$

From this and equation (1), $a=-8, b=9$.

Thus, the solution we are looking for is:

$$
\frac{6 x+1}{6 x^{2}+19 x+15}=\frac{-8}{x+\frac{3}{2}}+\frac{9}{x+\frac{5}{3}}=\frac{27}{3 x+5}-\frac{16}{2 x+3}
$$