## Problem Statement

Calculate the indefinite integral:

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

$$
\int \frac{3 x^{3}+2 x^{2}+1}{(x+2)(x-2)(x-1)} d x=\int \frac{3 x^{3}+2 x^{2}+1}{x^{3}-x^{2}-4 x+4} d x=
$$

Under the integral, there is an improper fraction. Let's separate the integer part:

$$
\begin{array}{ll}
3 x^{3}+2 x^{2}+1 & \mid x^{3}-x^{2}-4 x+4 \\
\frac{3 x^{3}-3 x^{2}-12 x+12}{5 x^{2}+12 x-11} &
\end{array}
$$

We get:

$$
=\int\left(3+\frac{5 x^{2}+12 x-11}{x^{3}-x^{2}-4 x+4}\right) d x=\int 3 d x+\int \frac{5 x^{2}+12 x-11}{(x+2)(x-2)(x-1)} d x=
$$

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

$$
\begin{aligned}
& \frac{5 x^{2}+12 x-11}{(x+2)(x-2)(x-1)}=\frac{A}{x+2}+\frac{B}{x-2}+\frac{C}{x-1}= \\
& =\frac{A(x-2)(x-1)+B(x+2)(x-1)+C(x+2)(x-2)}{(x+2)(x-2)(x-1)}= \\
& =\frac{A\left(x^{2}-3 x+2\right)+B\left(x^{2}+x-2\right)+C\left(x^{2}-4\right)}{(x+2)(x-2)(x-1)}= \\
& =\frac{(A+B+C) x^{2}+(-3 A+B) x+(2 A-2 B-4 C)}{(x+2)(x-2)(x-1)}= \\
& \left\{\begin{array}{l}
A+B+C=5 \\
-3 A+B=12 \\
2 A-2 B-4 C=-11
\end{array}\right.
\end{aligned}
$$

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

$$
\left\{\begin{array}{l}
A+B+C=5 \\
-3 A+B=12 \\
6 A+2 B=9
\end{array}\right.
$$

Add to the third row the second row multiplied by 2:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B+C=5 \\
-3 A+B=12 \\
4 B=33
\end{array}\right. \\
& \left\{\begin{array}{l}
A+B+C=5 \\
-3 A+B=12 \\
B=\frac{33}{4}
\end{array}\right. \\
& \left\{\begin{array}{l}
A+C=-\frac{13}{4} \\
A=-\frac{5}{4} \\
B=\frac{33}{4}
\end{array}\right. \\
& \left\{\begin{array}{l}
C=-2 \\
A=-\frac{5}{4} \\
B=\frac{33}{4}
\end{array}\right. \\
& \frac{5 x^{2}+12 x-11}{(x+2)(x-2)(x-1)}=-\frac{5}{4(x+2)}+\frac{33}{4(x-2)}-\frac{2}{x-1}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int 3 d x+\int\left(-\frac{5}{4(x+2)}+\frac{33}{4(x-2)}-\frac{2}{x-1}\right) d x= \\
& =3 x-\frac{5}{4} \cdot \ln |x+2|+\frac{33}{4} \cdot \ln |x-2|-2 \cdot \ln |x-1|+C
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-8$ »

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

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## Problem Kuznetsov Integrals $5-9$

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