Example 2. Decompose into partial fractions: $\frac{1}{(x+1)(x+2)^{2}(x+3)^{3}}$. Solution. According to Theorem 2, we have the decomposition

$$
\begin{aligned}
& \frac{1}{(x+1)(x+2)^{2}(x+3)^{3}}= \\
& \quad=\frac{A}{x+1}+\frac{B}{x+2}+\frac{C}{(x+2)^{2}}+\frac{D}{(x+3)}+\frac{E}{(x+3)^{2}}+\frac{F}{(x+3)^{3}}
\end{aligned}
$$

We bring the right-hand side of this equation to a common denominator. According to Theorem 3, we arrive at the equation

$$
\begin{aligned}
& 1=A(x+2)^{2}(x+3)^{3}+B(x+2)(x+1)(x+3)^{3}+ \\
& \quad+C(x+1)(x+3)^{3}+D(x+1)(x+2)^{2}(x+3)^{2}+ \\
& \quad+E(x+1)(x+2)^{2}(x+3)+F(x+2)^{2}(x+1)
\end{aligned}
$$

We will find the unknown coefficients by combining Theorems 4 and 5. First, we act according to Theorem 5.

1) Substitute $x=-2$ into the obtained equation. We get the equation $1=-C$ (all terms on the right side, except one, become zero). Hence, $C=-1$.
2) Now substitute $x=-1: 1=8 A$. We get $A=\frac{1}{8}$.
3) Substitute $x=-3, -2 F=1, F=-\frac{1}{2}$.
4) Substitute $x=0: 1=108 A+54 B+27 C+36 D+12 E+4 F \Rightarrow$ $\Rightarrow 54 B+36 D+12 E=16.5$. This equation will be used below.
To determine the remaining coefficients (taking into account the already obtained ones), we use Theorem 4.
5) We equate the coefficients of both sides for $x^{5}$ and $x^{4}$:

$$
\begin{aligned}
& \qquad x^{5}: 0=A+B+D \Rightarrow B+D=-\frac{1}{8} . \\
& x^{4}: 0=13 A+12 B+C+11 D+E \Rightarrow 12 B+11 D+E=-\frac{5}{8} . \\
& \text { 6) From the system }\left\{\begin{array}{l}
54 B+36 D+12 E=16.5, \\
B+D=-\frac{1}{8}, \\
12 B+11 D+E=-\frac{5}{8}
\end{array} \quad \text { we find: } E=-\frac{5}{4},\right. \\
& D=-\frac{17}{8}, B=2 . \\
& \text { Ans. } \\
& \frac{1}{(x+1)(x+2)^{2}(x+3)^{3}}= \\
& =\frac{1}{8(x+1)}+\frac{2}{x+2}-\frac{1}{(x+2)^{2}}-\frac{17}{8(x+3)}-\frac{5}{4(x+3)^{2}}-\frac{1}{2(x+3)^{3}} .
\end{aligned}
$$

Note. In further examples, we omit the expansion of brackets in the numerator of the right fraction. The search for coefficients for certain powers of the variable is usually performed mentally. Example 3. Decompose into partial fractions: $\frac{2 x^{4}-13 x^{3}+32 x^{2}-24 x+1}{x^{3}-5 x^{2}+6 x}$.
Solution.

1) The given fraction is improper, so we first separate the integer part.

$$
\begin{aligned}
& 2 x^{4}-13 x^{3}+32 x^{2}-24 x+1 \left\lvert\, \frac{x^{3}-5 x^{2}+6 x}{2 x-3}\right. \\
& \frac{2 x^{4}-10 x^{3}+12 x^{2}}{-3 x^{3}+20 x^{2}-24 x+1} \\
& \quad \frac{-3 x^{3}+15 x^{2}-18 x}{5 x^{2}-6 x+1}
\end{aligned}
$$

We get $\frac{2 x^{4}-13 x^{3}+32 x^{2}-24 x+1}{x^{3}-5 x^{2}+6 x}=2 x-3+\frac{5 x^{2}-6 x+1}{x^{3}-5 x^{2}+6 x}$.

2) It remains to decompose the fraction on the right side of this equation into simpler fractions, for which we first factor the denominator of this fraction into simple factors.

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

We equate the coefficients of the same powers of the numerators in the right and left fractions. This leads to a linear system of three equations with three unknowns $A, B$, and $C$.

$$
\left\{\begin{array} { l } 
{ x ^ { 2 } : A + B + C = 5 } \\
{ x ^ { 1 } : - 5 A - 2 B - 3 C = - 6 , } \\
{ x ^ { 0 } : 6 A = 1 \Rightarrow A = 1 / 6 }
\end{array} \Rightarrow \left\{\begin{array} { l } 
{ B + C = \frac { 2 9 } { 6 } , } \\
{ 2 B + 3 C = \frac { 3 1 } { 6 } }
\end{array} \Rightarrow \left\{\begin{array}{l}
A=\frac{1}{6} \\
B=\frac{28}{3} \\
C=-\frac{9}{2}
\end{array}\right.\right.\right.
$$

Answer. $\frac{2 x^{4}-13 x^{3}+32 x^{2}-24 x+1}{x^{3}-5 x^{2}+6 x}=2 x-3+\frac{1}{6 x}+\frac{28}{3(x-3)}-\frac{9}{2(x-2)}$.

Further examples with brief solutions, please solve on your own.