## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{5}+2 x^{4}-2 x^{3}+5 x^{2}-7 x+9}{(x+3)(x-1) x} d x
$$
## Solution

$$
\int \frac{x^{5}+2 x^{4}-2 x^{3}+5 x^{2}-7 x+9}{(x+3)(x-1) x} d x=\int \frac{x^{5}+2 x^{4}-2 x^{3}+5 x^{2}-7 x+9}{x^{3}+2 x^{2}-3 x} d x=
$$

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

$$
\begin{array}{cc}
x^{5}+2 x^{4}-2 x^{3}+5 x^{2}-7 x+9 & \frac{\mid x^{3}+2 x^{2}-3 x}{x^{2}+1} \\
\frac{x^{5}+2 x^{4}-3 x^{3}}{x^{3}+5 x^{2}-7 x+9} \\
\frac{x^{3}+2 x^{2}-3 x}{3 x^{2}-4 x+9} &
\end{array}
$$

We get:

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

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

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

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

Add the first row to the second:

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

Then we get:

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

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}+5-22 »

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

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## Problem Kuznetsov Integrals $5-23$

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