## Problem Statement

Calculate the definite integral:

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

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

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 $\% \mathrm{D} 1 \% 82 \% \mathrm{D} 0 \% \mathrm{~B} 5 \% \mathrm{D} 0 \% \mathrm{~B} 3 \% \mathrm{D} 1 \% 80 \% \mathrm{D} 0 \% \mathrm{~B} 0 \% \mathrm{D} 0 \% \mathrm{BB} \% \mathrm{D} 1 \% 8 \mathrm{~B} 4-3$ » Categories: Kuznetsov's Problem Book Integrals Problem 4 | Integrals

- Last edited: 23:10, March 4, 2009.
- Content is available under CC-BY-SA 3.0.

Created by GeeTeatoo

## Problem Kuznetsov Integrals 4-4

## Material from PlusPi