Is there an analogous rule of integration for the quotient rule?
For a rational function $P(x)/Q(x)$, the answer is affirmative, and it's called the Ostrogradsky-Hermite method. I adapt this answer of mine.
To integrate a rational function $P(x)/Q(x)$ without decomposing it into partial fractions and without finding the roots of the denominator, we can use the Ostrogradski-Hermite method. You can find a description of this method in section 2.1 of Gradshteyn and Ryzhik's Table of Integrals, Series, and Products, where the identity $(2)$ bellow is given. The formula $(1)$ appears also on the Ostrogradsky's Wikipedia page.
Assume that $\deg P(x)<\deg $ $Q(x)$. There exist polynomials $P_{1}(x)$, $P_{2}(x)$, $Q_{1}(x)$ and $Q_{2}(x)$, with $Q_{1}(x)=\gcd \left\{ Q(x), Q^{\prime }(x)\right\}$ and $Q_{2}(x)=Q(x)/Q_{1}(x)$, $\deg P_{1}(x)<\deg Q_{1}(x)$, $\deg P_{2}(x)<\deg Q_{2}(x)$, such that
\begin{equation}
\int \frac{P(x)}{Q(x)}dx=\frac{P_{1}(x)}{Q_{1}(x)}+\int \frac{P_{2}(x)}{
Q_{2}(x)}dx.\tag{1}
\end{equation}
Indeed, by differentiation and multiplication by $P(x)$, we have
\begin{eqnarray*}
P(x) &=&\frac{P_{1}^{\prime }(x)Q_{1}(x)-P_{1}(x)Q_{1}^{\prime }(x)}{\left\{
Q_{1}(x)\right\} ^{2}}Q(x)+\frac{P_{2}(x)}{Q_{2}(x)}Q(x) \\
&=&P_{1}^{\prime }(x)\frac{Q(x)}{Q_{1}(x)}-P_{1}(x)\frac{Q_{1}^{\prime }(x)}{Q_{1}(x)}\frac{Q(x)}{Q_{1}(x)}+P_{2}(x)\frac{Q(x)}{Q_{2}(x)} \\
&=&P_{1}^{\prime }(x)Q_{2}(x)-P_{1}(x)\left\{ \frac{Q_{1}^{\prime }(x)}{Q_{1}(x)}Q_{2}(x)\right\} +P_{2}(x)Q_{1}(x)
\end{eqnarray*}
or
\begin{equation}
P(x)=P_{1}^{\prime }(x)Q_{2}(x)-P_{1}(x)\left\{ T(x)-Q_{2}^{\prime }(x)\right\}+P_{2}(x)Q_{1}(x),\tag{2}
\end{equation}
with $T(x)=Q^{\prime }(x)/Q_{1}(x)$, because from
\begin{equation*}
Q^{\prime }(x)=\left\{ Q_{1}(x)Q_{2}(x)\right\} ^{\prime }=Q_{1}^{\prime
}(x)Q_{2}(x)+Q_{1}(x)Q_{2}^{\prime }(x)=T(x)Q_{1}(x)
\end{equation*}
we obtain
\begin{equation*}
\frac{Q_{1}^{\prime }(x)}{Q_{1}(x)}Q_{2}(x)+Q_{2}^{\prime }(x)=T(x).
\end{equation*}
To find the coefficients of the polynomials $P_{1}(x)$ and $P_{2}(x)$ we equate the coefficients of like powers of $x$ and/or make $x=x_1,x_2,\dots$ until we obtain a system of linear equations in those coefficients.
