## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{x^{3}-17}{x^{2}-4 x+3} d x
$$
## Solution

$$
\int \frac{x^{3}-17}{x^{2}-4 x+3} d x=
$$

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

$$
\begin{aligned}
& x^{3}-17 \\
& x^{2}-4 x+3 \\
& \frac{x^{3}-4 x^{2}+3 x}{4 x^{2}-3 x}-17 \\
& \frac{4 x^{2}-16 x+12}{13 x-29}
\end{aligned}
$$

We get:

$$
=\int\left(x+4+\frac{13 x-29}{x^{2}-4 x+3}\right) d x=\int(x+4) d x+\int \frac{13 x-29}{(x-1)(x-3)} d x=
$$

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

$$
\begin{aligned}
& \frac{13 x-29}{(x-1)(x-3)}=\frac{A}{x-1}+\frac{B}{x-3}=\frac{A(x-3)+B(x-1)}{(x-1)(x-3)}=\frac{(A+B) x-(3 A+B)}{(x-1)(x-3)} \\
& \left\{\begin{array}{l}
A+B=13 \\
-(3 A+B)=-29
\end{array}\right. \\
& \left\{\begin{array}{l}
A=8 \\
B=5
\end{array}\right. \\
& \frac{x+1}{x(x-1)}=\frac{8}{x-1}+\frac{5}{x-3}
\end{aligned}
$$

Then we get:

$$
=\int(x+4) d x+\int\left(\frac{8}{x-1}+\frac{5}{x-3}\right) d x=\frac{x^{2}}{2}+4 x+8 \ln |x-1|+5 \ln |x-3|+C
$$

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\%D1\%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}+5-3 »

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

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