## Problem Statement

Calculate the definite integral:

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

$$
\begin{aligned}
& \int_{0}^{\sqrt{3}} \frac{\operatorname{arctg} x + x}{1 + x^{2}} d x = \int_{0}^{\sqrt{3}} \frac{\operatorname{arctg} x}{1 + x^{2}} d x + \int_{0}^{\sqrt{3}} \frac{x}{1 + x^{2}} d x = \\
& = \frac{1}{2} \int_{0}^{\sqrt{3}} \operatorname{arctg} x d(\operatorname{arctg} x) + \frac{1}{2} \int_{0}^{\sqrt{3}} \frac{d\left(1 + x^{2}\right)}{1 + x^{2}} = \left. \frac{1}{2} \left( \operatorname{arctg}^{2} x + \ln \left(1 + x^{2}\right) \right) \right|_{0}^{\sqrt{3}} = \\
& = \frac{1}{2} \left( \operatorname{arctg}^{2} \sqrt{3} + \ln \left(1 + (\sqrt{3})^{2}\right) \right) - \frac{1}{2} \left( \operatorname{arctg}^{2} 0 + \ln \left(1 + 0^{2}\right) \right) = \\
& = \frac{1}{2} \cdot \left( \left( \frac{\pi}{3} \right)^{2} + \ln 4 \right) - \frac{1}{2} \left( 0^{2} + \ln 1 \right) = \frac{1}{2} \cdot \left( \frac{\pi^{2}}{9} + 2 \ln 2 \right) - \frac{1}{2} \cdot (0 + 0) = \\
& = \frac{\pi^{2}}{18} + \ln 2
\end{aligned}
$$

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}+4-12 »

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

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## Problem Kuznetsov Integrals 4-13

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