## Problem Statement

Calculate the indefinite integral:

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

$$
\begin{aligned}
& \int \frac{8 x - \operatorname{arctg} 2 x}{1 + 4 x^{2}} d x = \int \frac{8 x}{1 + 4 x^{2}} d x - \int \frac{\operatorname{arctg} 2 x}{1 + 4 x^{2}} d x = \\
& = \int \frac{d\left(1 + 4 x^{2}\right)}{1 + 4 x^{2}} - \frac{1}{2} \cdot \int \operatorname{arctg} 2 x \cdot d(\operatorname{arctg} 2 x) = \ln \left|1 + 4 x^{2}\right| - \frac{1}{4} \cdot \operatorname{arctg}^{2} 2 x + C
\end{aligned}
$$

Source — "http://pluspi.org/wiki/index.php/\�\�\�\�\�\�\�\�\�\�\�\�

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Categories: Kuznetsov Problem Book Integrals Problem $3 \mid$ Integrals

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## Problem Kuznetsov Integrals 3-23

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