## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}+2 x^{2}+3}{(x-1)(x-2)(x-3)} d x
$$
## Solution

$$
\int \frac{x^{3}+2 x^{2}+3}{(x-1)(x-2)(x-3)} d x=\int \frac{x^{3}+2 x^{2}+3}{x^{3}-6 x^{2}+11 x-6} d x=
$$

Under the integral, we have an improper fraction. Let's separate the integer part:

$$
\begin{array}{ll}
x^{3}+2 x^{2}+3 & \mid x^{3}-6 x^{2}+11 x-6 \\
\frac{x^{3}-6 x^{2}+11 x-6}{8 x^{2}-11 x+9} & 1
\end{array}
$$

We get:

$$
=\int\left(1+\frac{8 x^{2}-11 x+9}{x^{3}-6 x^{2}+11 x-6}\right) d x=\int d x+\int \frac{8 x^{2}-11 x+9}{(x-1)(x-2)(x-3)} d x=
$$

Let's decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{8 x^{2}-11 x+9}{(x-1)(x-2)(x-3)}=\frac{A}{x-1}+\frac{B}{x-2}+\frac{C}{x-3}= \\
& =\frac{A(x-2)(x-3)+B(x-1)(x-3)+C(x-1)(x-2)}{(x-1)(x-2)(x-3)}= \\
& =\frac{A\left(x^{2}-5 x+6\right)+B\left(x^{2}-4 x+3\right)+C\left(x^{2}-3 x+2\right)}{(x-1)(x-2)(x-3)}= \\
& =\frac{(A+B+C) x^{2}+(-5 A-4 B-3 C) x+(6 A+3 B+2 C)}{(x-1)(x-2)(x-3)}= \\
& \left\{\begin{array}{l}
A+B+C=8 \\
-5 A-4 B-3 C=-11 \\
6 A+3 B+2 C=9
\end{array}\right.
\end{aligned}
$$

Add the second row to the third:

$$
\left\{\begin{array}{l}
A+B+C=8 \\
-5 A-4 B-3 C=-11 \\
A-B-C=-2
\end{array}\right.
$$

Add the first row to the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B+C=8 \\
-5 A-4 B-3 C=-11 \\
2 A=6
\end{array}\right. \\
& \left\{\begin{array}{l}
B+C=5 \\
-4 B-3 C=4 \\
A=3
\end{array}\right.
\end{aligned}
$$

Add the first row multiplied by 4 to the second row:

$$
\begin{aligned}
& \left\{\begin{array}{l}
B+C=5 \\
C=24 \\
A=3
\end{array}\right. \\
& \left\{\begin{array}{l}
B=-19 \\
C=24 \\
A=3
\end{array}\right. \\
& \frac{8 x^{2}-11 x+9}{(x-1)(x-2)(x-3)}=\frac{3}{x-1}-\frac{19}{x-2}+\frac{24}{x-3}
\end{aligned}
$$

Then we get:

$$
=\int d x+\int\left(\frac{3}{x-1}-\frac{19}{x-2}+\frac{24}{x-3}\right) d x=x+3 \cdot \ln |x-1|-19 \cdot \ln |x-2|+24 \cdot \ln |x-3|+C
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_\�\� \%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD\%D1\%82\%D0\%B5\%D0\%B3 $\% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} \_5-7$ »

Categories: Kuznetsov's Problem Book Integrals Problem 5 | Integrals

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## Problem Kuznetsov Integrals 5-8

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