## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{2} \frac{x^{3}}{x^{2}+4} d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{2} \frac{x^{3}}{x^{2}+4} d x=\frac{1}{2} \cdot \int_{0}^{2} \frac{x^{2}}{x^{2}+4} d\left(x^{2}\right)=\frac{1}{2} \cdot \int_{0}^{2} \frac{x^{2}+4-4}{x^{2}+4} d\left(x^{2}\right)= \\
& =\frac{1}{2} \cdot \int_{0}^{2} d\left(x^{2}\right)-\frac{1}{2} \cdot \int_{0}^{2} \frac{4}{x^{2}+4} d\left(x^{2}\right)=\left.\frac{1}{2} \cdot x^{2}\right|_{0} ^{2}-\left.2 \ln \left(x^{2}+4\right)\right|_{0} ^{2}= \\
& =\frac{1}{2} \cdot 2^{2}-\frac{1}{2} \cdot 0^{2}-2 \ln \left(2^{2}+4\right)+2 \ln \left(0^{2}+4\right)= \\
& =2-0-2 \ln 8+2 \ln 4=2-2 \ln 2^{3}+2 \ln 2^{2}=2-6 \ln 2+4 \ln 2= \\
& =2-2 \ln 2
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} 4-4$ "

Categories: Kuznetsov's Problem Book Integrals Problem 4 | Integrals

- Last edited: March 3, 2009, at 16:42.
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## Problem Kuznetsov Integrals 4-5

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