## Problem Statement

Calculate the indefinite integral:

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

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

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

$$
\begin{array}{ll}
4 x^{3}+x^{2}+2 & \mid x^{3}-3 x^{2}+2 x \\
\frac{4 x^{3}-12 x^{2}+8 x}{13 x^{2}-8 x+2} & 4
\end{array}
$$

We get:

$$
=\int\left(4+\frac{13 x^{2}-8 x+2}{x^{3}-3 x^{2}+2 x}\right) d x=\int 4 d x+\int \frac{13 x^{2}-8 x+2}{x(x-1)(x-2)} d x=
$$

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

$$
\begin{aligned}
& \frac{13 x^{2}-8 x+2}{x(x-1)(x-2)}=\frac{A}{x}+\frac{B}{x-1}+\frac{C}{x-2}= \\
& =\frac{A(x-1)(x-2)+B x(x-2)+C x(x-1)}{x(x-1)(x-2)}= \\
& =\frac{A\left(x^{2}-3 x+2\right)+B\left(x^{2}-2 x\right)+C\left(x^{2}-x\right)}{x(x-1)(x-2)}= \\
& =\frac{(A+B+C) x^{2}+(-3 A-2 B-C) x+2 A}{x(x-1)(x-2)}= \\
& \left\{\begin{array}{l}
A+B+C=13 \\
-3 A-2 B-C=-8 \\
2 A=2
\end{array}\right.
\end{aligned}
$$

$$
\left\{\begin{array}{l}
B+C=12 \\
-2 B-C=-5 \\
A=1
\end{array}\right.
$$

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

$$
\begin{aligned}
& \left\{\begin{array}{l}
B+C=12 \\
C=19 \\
A=1
\end{array}\right. \\
& \left\{\begin{array}{l}
B=-7 \\
C=19 \\
A=1
\end{array}\right. \\
& \frac{13 x^{2}-8 x+2}{x(x-1)(x-2)}=\frac{1}{x}-\frac{7}{x-1}+\frac{19}{x-2}
\end{aligned}
$$

Then we get:

$$
\begin{aligned}
& =\int 4 d x+\int\left(\frac{1}{x}-\frac{7}{x-1}+\frac{19}{x-2}\right) d x= \\
& =4 x+\ln |x|-7 \cdot \ln |x-1|+19 \cdot \ln |x-2|+C
\end{aligned}
$$

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}+5-12 »

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

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