   \textbf{4.} Spo�t�te ur�it� integr�l
    $$
        \int_{-1}^{1} \frac{x}{x^2+x+1} \dx.
    $$

    \textbf{�e�en�}
    \begin{multline*}
        \int_{-1}^{1} \frac{x}{x^2+x+1} \dx =
        \int_{-1}^{1} \left(\frac{x + \frac12}{x^2 + x + 1} - \frac{\frac12}{x^2 + x + 1}\right) \dx =\\
        =\left[\frac12 \log(x^2+x+1) - \frac12 \cdot \frac{2}{\sqrt3}\arctg\left(\frac{2x}{\sqrt3} +
            \frac{1}{\sqrt3}\right)\right]_{-1}^{1} =\\
        =\left[\frac12 \log(x^2+x+1) + \frac1{\sqrt3} \arctg\left(\frac{2x+1}{\sqrt 3}\right) \right]_{-1}^{1} =\\
        =\frac12 \log 3 + \frac1{\sqrt3} \arctg \sqrt3 - \frac12 \log 1 - \frac1{\sqrt3} \arctg \frac{1}{\sqrt3} =\\
        =\frac12 \log 3 + \frac{\pi}{3\sqrt3} - \frac{\pi}{6\sqrt3} = \frac12 \log 3 + \frac{\pi}{6\sqrt3}.
    \end{multline*}
