## Problem Statement

Calculate the indefinite integral:

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

$$
\begin{aligned}
& \int \frac{x^{3}-3 x^{2}-12}{(x-4)(x-3) x} d x=\int \frac{x^{3}-3 x^{2}-12}{x^{3}-7 x^{2}+12 x} d x= \\
& =\int\left(1+\frac{4 x^{2}-12 x-12}{x^{3}-7 x^{2}+12 x}\right) d x=\int d x+\int \frac{4 x^{2}-12 x-12}{(x-4)(x-3) x} d x=
\end{aligned}
$$

We decompose the proper rational fraction into partial fractions using the method of undetermined coefficients:

$$
\begin{aligned}
& \frac{4 x^{2}-12 x-12}{(x-4)(x-3) x}=\frac{A}{x-4}+\frac{B}{x-3}+\frac{C}{x}= \\
& =\frac{A(x-3) x+B(x-4) x+C(x-4)(x-3)}{(x-4)(x-3) x}= \\
& =\frac{A\left(x^{2}-3 x\right)+B\left(x^{2}-4 x\right)+C\left(x^{2}-7 x+12\right)}{(x-4)(x-3) x}= \\
& =\frac{(A+B+C) x^{2}+(-3 A-4 B-7 C) x+12 C}{(x-4)(x-3) x}= \\
& \left\{\begin{array}{l}
A+B+C=4 \\
-3 A-4 B-7 C=-12 \\
12 C=-12
\end{array}\right. \\
& \left\{\begin{array}{l}
A+B=5 \\
-3 A-4 B=-19 \\
C=-1
\end{array}\right.
\end{aligned}
$$

Adding the first row multiplied by 4 to the second row:

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

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

Then we get:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+5-11$ »

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

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## Problem Kuznetsov Integrals 5-12

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