## Problem Statement

Calculate the indefinite integral:

$$
\int \frac{3 x^{3}+25}{x^{2}+3 x+2} d x
$$
## Solution

$$
\int \frac{3 x^{3}+25}{x^{2}+3 x+2} d x=
$$

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

$$
\begin{array}{ll}
\frac{3 x^{3}+25}{3 x^{3}+9 x^{2}+6 x} & \left\lvert\, \frac{x^{2}+3 x+2}{3 x-9}\right. \\
-9 x^{2}-6 x+25 & \\
\frac{-9 x^{2}-27 x-18}{21 x+43} &
\end{array}
$$

We get:

$$
=\int\left(3 x-9+\frac{21 x+43}{x^{2}+3 x+2}\right) d x=\int(3 x-9) d x+\int \frac{21 x+43}{(x+1)(x+2)} d x=
$$

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

$$
\begin{aligned}
& \frac{21 x+43}{(x+1)(x+2)}=\frac{A}{x+1}+\frac{B}{x+2}=\frac{A(x+2)+B(x+1)}{(x+1)(x+2)}=\frac{(A+B) x+(2 A+B)}{(x+1)(x+2)} \\
& \left\{\begin{array}{l}
A+B=21 \\
2 A+B=43
\end{array}\right. \\
& \left\{\begin{array}{l}
A=22 \\
B=-1
\end{array}\right. \\
& \frac{21 x+43}{(x+1)(x+2)}=\frac{22}{x+1}-\frac{1}{x+2}
\end{aligned}
$$

Then we get:

$$
=\int(3 x-9) d x+\int\left(\frac{22}{x+1}-\frac{1}{x+2}\right) d x=\frac{3 x^{2}}{2}-9 x+22 \cdot \ln |x+1|-\ln |x+2|+C
$$

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%D0%B5%D0%B3%D1%80%D0%B0%D0%BB%D1%8B_%D0%97%D0%B0%D0%B4%D0%B0%D1%87%D0%B0_5"

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

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## Problem Kuznetsov Integrals 5-7

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