   \textbf{4.} Spo�t�te integr�l
      $$
            \int_{a}^{+\infty} \frac{dx}{x^2} \quad (a>0).
      $$

      \textbf{�e�en�} Zn�me primitivn� funkci k funkci $\frac1{x^2}$, tedy m��eme ps�t
      $$
            \int_{a}^{+\infty} \frac{dx}{x^2} =
            \left[ -\frac1x \right]_{a}^{+\infty} =
            \lim_{x\to\infty}\left(-\frac 1x \right) - \lim_{x\to a_+}\left(-\frac1x\right) =
            0 - \frac 1a = \frac 1a.
      $$
