## Problem Statement

Calculate the definite integral:

$$
\int_{0}^{\frac{1}{2}} \frac{8 x-\operatorname{arctg} 2 x}{1+4 x^{2}} d x
$$
## Solution

$$
\begin{aligned}
& \int_{0}^{\frac{1}{2}} \frac{8 x-\operatorname{arctg} 2 x}{1+4 x^{2}} d x=\int_{0}^{\frac{1}{2}} \frac{8 x}{1+4 x^{2}} d x-\int_{0}^{\frac{1}{2}} \frac{\operatorname{arctg} 2 x}{1+4 x^{2}} d x= \\
& =\int_{0}^{\frac{1}{2}} \frac{d\left(1+4 x^{2}\right)}{1+4 x^{2}}-\frac{1}{2} \cdot \int_{0}^{\frac{1}{2}} \operatorname{arctg} 2 x \cdot d(\operatorname{arctg} 2 x)=\left.\ln \left(1+4 x^{2}\right)\right|_{0} ^{\frac{1}{2}}-\left.\frac{1}{4} \cdot \operatorname{arctg}^{2} 2 x\right|_{0} ^{\frac{1}{2}}= \\
& =\ln \left(1+4 \cdot\left(\frac{1}{2}\right)^{2}\right)-\ln \left(1+4 \cdot 0^{2}\right)-\frac{1}{4} \cdot \operatorname{arctg}^{2}\left(2 \cdot \frac{1}{2}\right)+\frac{1}{4} \cdot \operatorname{arctg}^{2}(2 \cdot 0)= \\
& =\ln 2-\ln 1-\frac{1}{4} \cdot \operatorname{arctg}^{2} 1+\frac{1}{4} \cdot \operatorname{arctg}^{2} 0=\ln 2-0-\frac{1}{4} \cdot\left(\frac{\pi}{4}\right)^{2}+\frac{1}{4} \cdot 0^{2}= \\
& =\ln 2-\frac{\pi^{2}}{64}
\end{aligned}
$$

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B}+4-7$ >

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

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## Problem Kuznetsov Integrals 4-8

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