## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-3)(x-2)} d x
$$
## Solution

$$
\int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-3)(x-2)} d x=\int \frac{x^{3}-3 x^{2}-12}{x^{3}-9 x^{2}+26 x-24} d x=
$$

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

$$
\begin{array}{ll}
x^{3}-3 x^{2}-12 & \mid x^{3}-9 x^{2}+26 x-24 \\
\frac{x^{3}-9 x^{2}+26 x-24}{6 x^{2}-26 x+12} & 1
\end{array}
$$

We get:

$$
=\int\left(1+\frac{6 x^{2}-26 x+12}{x^{3}-9 x^{2}+26 x-24}\right) d x=\int d x+\int \frac{6 x^{2}-26 x+12}{(x-4)(x-3)(x-2)} d x=
$$

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

$$
\begin{aligned}
& \frac{6 x^{2}-26 x+12}{(x-4)(x-3)(x-2)}=\frac{A}{x-4}+\frac{B}{x-3}+\frac{C}{x-2}= \\
& =\frac{A(x-3)(x-2)+B(x-4)(x-2)+C(x-4)(x-3)}{(x-4)(x-3)(x-2)}= \\
& =\frac{A\left(x^{2}-5 x+6\right)+B\left(x^{2}-6 x+8\right)+C\left(x^{2}-7 x+12\right)}{(x-4)(x-3)(x-2)}= \\
& =\frac{(A+B+C) x^{2}+(-5 A-6 B-7 C) x+(6 A+8 B+12 C)}{(x-4)(x-3)(x-2)}= \\
& \left\{\begin{array}{l}
A+B+C=6 \\
-5 A-6 B-7 C=-26 \\
6 A+8 B+12 C=12
\end{array}\right.
\end{aligned}
$$

Add the first row multiplied by -6 to the third row:

$$
\left\{\begin{array}{l}
A+B+C=6 \\
-5 A-6 B-7 C=-26 \\
2 B+6 C=-24
\end{array}\right.
$$

Add the first row multiplied by 5 to the second row:

$$
\left\{\begin{array}{l}
A+B+C=6 \\
-B-2 C=4 \\
2 B+6 C=-24
\end{array}\right.
$$

Add the second row multiplied by 3 to the third row:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B+C=6 \\
-B-2 C=4 \\
-B=-12
\end{array}\right. \\
& \left\{\begin{array}{l}
A+C=-6 \\
-2 C=16 \\
B=12
\end{array}\right. \\
& \left\{\begin{array}{l}
A=2 \\
C=-8 \\
B=12
\end{array}\right. \\
& \frac{6 x^{2}-26 x+12}{(x-4)(x-3)(x-2)}=\frac{2}{x-4}+\frac{12}{x-3}-\frac{8}{x-2}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int d x+\int\left(\frac{2}{x-4}+\frac{12}{x-3}-\frac{8}{x-2}\right) d x= \\
& =x+2 \cdot \ln |x-4|+12 \cdot \ln |x-3|-8 \cdot \ln |x-2|+C
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\� \%D0\%9A\%D1\%83\%D0\%B7\%D0\%BD\%D0\%B5\%D1\%86\%D0\%BE\%D0\%B2_\%D0\%98\%D0\%BD $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-10$ » Categories: Kuznetsov's Problem Book Integrals Problem 5 | Integrals

Ukrainian Banner Network

- Last edited: 20:47, April 4, 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 5-11

## Material from Plusi