## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\sqrt{3}} \frac{x-(\operatorname{arctg} x)^{4}}{1+x^{2}} d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{\sqrt{3}} \frac{x-(\operatorname{arctg} x)^{4}}{1+x^{2}} d x=\int_{0}^{\sqrt{3}} \frac{x d x}{1+x^{2}}-\int_{0}^{\sqrt{3}} \frac{(\operatorname{arctg} x)^{4}}{1+x^{2}}= \\
& =\frac{1}{2} \int_{0}^{\sqrt{3}} \frac{d\left(1+x^{2}\right)}{1+x^{2}}-\int_{0}^{\sqrt{3}}(\operatorname{arctg} x)^{4} d(\operatorname{arctg} x)=\left.\frac{1}{2} \cdot \ln \left(1+x^{2}\right)\right|_{0} ^{\sqrt{3}}-\left.\frac{(\operatorname{arctg} x)^{5}}{5}\right|_{0} ^{\sqrt{3}}= \\
& =\frac{1}{2} \cdot \ln \left(1+(\sqrt{3})^{2}\right)-\frac{1}{2} \cdot \ln \left(1+0^{2}\right)-\frac{(\operatorname{arctg} \sqrt{3})^{5}}{5}+\frac{(\operatorname{arctg} 0)^{5}}{5}= \\
& =\frac{1}{2} \cdot \ln 4-\frac{1}{2} \cdot \ln 1-\frac{\left(\frac{\pi}{3}\right)^{5}}{5}+\frac{0^{5}}{5}=\ln 2-0-\frac{\left(\frac{\pi}{3}\right)^{5}}{5}+0=\ln 2-\frac{\pi^{5}}{5 \cdot 3^{5}}
\end{aligned}
$$

Source — «http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�_ $\% \mathrm{D} 0 \% 9 \mathrm{~A} \% \mathrm{D} 1 \% 83 \% \mathrm{D} 0 \% \mathrm{~B} 7 \% \mathrm{D} 0 \% \mathrm{BD} \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 1 \% 86 \% \mathrm{D} 0 \% \mathrm{BE} \% \mathrm{D} 0 \% \mathrm{~B} 2 \mathrm{H} 0 \mathrm{D} 0 \% 98 \% \mathrm{D} 0 \% \mathrm{BD}$ $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} 4-13$ "

Categories: Kuznetsov Problem Book Integrals Problem 4 | Integrals

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## Problem Kuznetsov Integrals 4-14

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