Example 1.7. Find $\int \frac{6 d x}{x(x-1)(x-2)(x-3)}$.
Solution. The decomposition of the integrand into a sum of the simplest fractions has the form:

$$
\frac{6}{x(x-1)(x-2)(x-3)}=\frac{A}{x}+\frac{B}{(x-1)}+\frac{C}{(x-2)}+\frac{-D}{(x-3)}=
$$

The coefficients of the decomposition $A, B, C, D$ are found by equating the numerators in the right and left parts of the equality after bringing the fractions in the right part to a common denominator. Continue the equality:

$$
\begin{gathered}
=\frac{A(x-1)(x-2)(x-3)+B x(x-2)(x-3)+C x(x-1)(x-3)+D x(x-1)(x-2)}{x(x-1)(x-2)(x-3)} \Rightarrow \\
\Rightarrow A(x-1)(x-2)(x-3)+B x(x-2)(x-3)+C x(x-1)(x-3)+ \\
+D x(x-1)(x-2) x(x-1)(x-2)(x-3)=6 .
\end{gathered}
$$

Expanding the brackets in the left part of the written equality and combining like terms, we get:

$$
\begin{aligned}
x^{3}(A+B+C+D) & +x^{2}(-6 A-5 B-4 C-3 D)+ \\
& +x(11 A+6 B+3 C+2 D)-6 A=6
\end{aligned}
$$

Equating further the coefficients of the same powers of $x$ in the left and right parts of the equality, we get the following system of equations to determine the unknown coefficients of the decomposition:

$$
\left\{\begin{array} { l } 
{ A + B + C + D = 0 } \\
{ 6 A + 5 B + 4 C + 3 D = 0 } \\
{ 1 1 A + 6 B + 3 C + 2 D = 0 } \\
{ - 6 A = 6 }
\end{array} \Rightarrow \left\{\begin{array}{l}
B+C+D=1 \\
5 B+4 C+3 D=6 \\
6 B+3 C+2 D=11 \\
A=-1
\end{array}\right.\right.
$$

After subtracting the second equation from the third equation of the obtained system, we get:

$$
\left\{\begin{array}{l}
B+C+D=1 \\
B-C-D=5 \\
6 B+4 C+3 D=6 \\
A=-1
\end{array}\right.
$$

Adding the first and second equations of the obtained system, we find $B=3$. Substituting the obtained value into the first and third equations of the system, we arrive at a system of two linear equations for the unknown coefficients $C$ and $D$:

$$
\left\{\begin{array}{l}
C+D=-2 \\
4 C+3 D=-9
\end{array}\right.
$$

From the first equation of this system, we have $C=-2-D$. Then, from the second equation of the system after substitution, we get:

$$
-4(2+D)+3 D=-9 \Rightarrow D=1
$$

Then, obviously, $C=-3$. Thus, the coefficients of the decomposition are:

$$
A=-1, B=3, C=-3, D=1
$$

This result could also have been obtained in a simpler way. If in the equation for determining the coefficients we set $x=0$, then we get $-6 A=6$, from which $A=-1$. Setting $x=1$ sequentially, we find $2 B=6$, from which $B=3$, for $x=2$ we get $-2 C=6$, from which $C=-3$, for $x=3$ we get $6 D=6$, from which $D=1$.

This method can always be used to find part of the coefficients of the decomposition of a rational fraction into a sum of the simplest fractions, if its denominator has real roots. The number of coefficients determined in this way is equal to the number of different real roots of the denominator. This allows reducing the number of linear equations in the system that needs to be solved to find the remaining coefficients of the decomposition. Note that when implementing the discussed method, complex roots of the denominator can also be used, but we will not dwell on this, as they are usually not used in the standard higher mathematics course in technical and technological universities.

Returning to the integral, we get

$$
\begin{aligned}
& \int \frac{6 d x}{x(x-1)(x-2)(x-3)}= \\
& =-\int \frac{d x}{x}+3 \int \frac{d x}{x-1}-3 \int \frac{d x}{x-2}+\int \frac{d x}{x-3}= \\
& =-\ln |x|+3 \ln |x-1|-3 \ln |x-2|+\ln |x-3|+C .
\end{aligned}
$$