## Problem Statement

Calculate the indefinite integral:

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

## Solution №1

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

## Solution №2

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

Under the integral, there is an improper fraction. Let's separate the integer part:

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

We get:

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+3-19$ »

Categories: Kuznetsov's Problem Book Integrals Problem 3 | Integrals

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## Problem Kuznetsov Integrals 3-20

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