Example 1. Find the integral $\int \frac{9-5 x}{x^{3}-6 x^{2}+11 x-6} d x$.
Solution. First, we decompose the proper rational fraction

$$
\frac{9-5 x}{x^{3}-6 x^{2}+11 x-6}
$$

into partial fractions, for which we need to find the roots of its denominator, i.e., the roots of the equation

$$
x^{3}-6 x^{2}+11 x-6=0 .
$$

One root is immediately apparent, which is $x_{1}=1$. Since

$$
\left(x^{3}-6 x^{2}+11 x-6\right):(x-1)=x^{2}-5 x+6
$$

then

$$
\begin{gathered}
x^{3}-6 x^{2}+11 x-6=(x-1)\left(x^{2}-5 x+6\right)= \\
=(x-1)(x-2)(x-3)
\end{gathered}
$$

therefore we also obtain

$$
x_{2}=2, x_{3}=3
$$

Thus, the polynomial has real simple roots

$$
x_{1}=1, x_{2}=2, x_{3}=3
$$

According to formula (4.29), we seek the decomposition of the given fraction into partial fractions

$$
\frac{9-5 x}{x^{3}-6 x^{2}+11 x-6}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}
$$

(The decomposition does not contain terms of the form $\frac{m x+n}{\left(x^{2}+p x+q\right)^{k}}$ because there are no complex roots.) Bringing the fractions on the right-hand side to a common denominator, we get

$$
\begin{gathered}
\frac{9-5 x}{x^{3}-6 x^{2}+11 x-6}= \\
=\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}
\end{gathered}
$$

We compare the numerators

$$
9-5 x=A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)
$$

Expanding the brackets in the right-hand side of equation (B) and grouping the terms, we find

$$
9-5 x=(A+B+C) x^{2}-(5 A+4 B+3 C) x+(6 A+3 B+2 C)
$$

Comparing the coefficients of like powers of $x$ in both sides of the equation, we obtain three equations for determining the unknown coefficients $A, B, C$ :

$$
\begin{gathered}
A+B+C=0 \\
5 A+4 B+3 C=5 \\
6 A+3 B+2 C=9
\end{gathered}
$$

Solving this system, we find: $A=2, B=1, C=-3$.

Therefore,

$$
\frac{9-5 x}{x^{3}-6 x^{2}+11 x-6}=\frac{2}{x-1}+\frac{1}{x-2}-\frac{3}{x-3}
$$

and thus

$$
\begin{gathered}
\int \frac{9-5 x}{x^{3}-6 x^{2}+11 x-6} d x=\int\left(\frac{2}{x-1}+\frac{1}{x-2}-\frac{3}{x-3}\right) d x= \\
=2 \int \frac{1}{x-1} d x+\int \frac{1}{x-2} d x-3 \int \frac{1}{x-3} d x=
\end{gathered}
$$

$$
\begin{gathered}
=2 \int \frac{d(x-1)}{x-1}+\int \frac{d(x-2)}{x-2}-3 \int \frac{d(x-3)}{x-3}= \\
=2 \ln |x-1|+\ln |x-2|-3 \ln |x-3|+C
\end{gathered}
$$

Thus,

$$
\int \frac{9-5 x}{x^{3}-6 x^{2}+11 x-6} d x=\ln \left|\frac{(x-1)^{2}(x-2)}{(x-3)^{3}}\right|+C
$$

Remark. The coefficients of the decomposition (A) could also be found by substituting $x=1, x=2, x=3$ into equation (B). Indeed, for $x=1$ we have $9-5=A(-1)(-2), 4=2 A$, from which $A=2$. Similarly, we obtain $B=1$, $C=-3$.