## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}-5 x^{2}+5 x+23}{(x-1)(x+1)(x-5)} d x
$$
## Solution

$$
\int \frac{x^{3}-5 x^{2}+5 x+23}{(x-1)(x+1)(x-5)} d x=\int \frac{x^{3}-5 x^{2}+5 x+23}{x^{3}-5 x^{2}-x+5} d x=
$$

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

$$
\begin{aligned}
& x^{3}-5 x^{2}+5 x+23 \quad \mid x^{3}-5 x^{2}-x+5 \\
& \frac{x^{3}-5 x^{2}-x+5}{6 x+18} 1
\end{aligned}
$$

We get:

$$
=\int\left(1+\frac{6 x+18}{x^{3}-5 x^{2}-x+5}\right) d x=\int d x+\int \frac{6 x+18}{(x-1)(x+1)(x-5)} d x=
$$

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

$$
\begin{aligned}
& \frac{6 x+18}{(x-1)(x+1)(x-5)}=\frac{A}{x-1}+\frac{B}{x+1}+\frac{C}{x-5}= \\
& =\frac{A(x+1)(x-5)+B(x-1)(x-5)+C(x-1)(x+1)}{(x-1)(x+1)(x-5)}= \\
& =\frac{A\left(x^{2}-4 x-5\right)+B\left(x^{2}-6 x+5\right)+C\left(x^{2}-1\right)}{(x-1)(x+1)(x-5)}= \\
& =\frac{(A+B+C) x^{2}+(-4 A-6 B) x+(-5 A+5 B-C)}{(x-1)(x+1)(x-5)} \\
& \left\{\begin{array}{l}
A+B+C=0 \\
-4 A-6 B=6 \\
-5 A+5 B-C=18
\end{array}\right.
\end{aligned}
$$

Add the first row to the third:

$$
\left\{\begin{array}{l}
A+B+C=0 \\
-4 A-6 B=6 \\
-4 A+6 B=18
\end{array}\right.
$$

Add the second row to the third:

$$
\begin{aligned}
& \left\{\begin{array}{l}
A+B+C=0 \\
-4 A-6 B=6 \\
-8 A=24
\end{array}\right. \\
& \left\{\begin{array}{l}
B+C=3 \\
-6 B=-6 \\
A=-3
\end{array}\right. \\
& \left\{\begin{array}{l}
C=2 \\
B=1 \\
A=-3
\end{array}\right. \\
& \frac{6 x+18}{(x-1)(x+1)(x-5)}=-\frac{3}{x-1}+\frac{1}{x+1}+\frac{2}{x-5}
\end{aligned}
$$

Then we get:

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

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}+5-21 »

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

- Last modified: 12:02, 6 April 2009.
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## Problem Kuznetsov Integrals $5-22$

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