\[\begin{array}{l}
 T17. \to \int {\frac{{dx}}{{{x^2} - 1}}}  = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| + C =  - {\mathop{\rm arcctgh}\nolimits} x + C \\ 
 {\rm{Dokaz}}\,{\rm{(Proof):}} \\ 
 \int {\frac{{dx}}{{{x^2} - 1}}}  = \frac{1}{2}\int {\frac{{(x + 1) - (x - 1)}}{{(x - 1)(x + 1)}}dx}  = \frac{1}{2}\int {\frac{{dx}}{{x - 1}}}  - \frac{1}{2}\int {\frac{{dx}}{{x + 1}}}  = \left| {\begin{array}{*{20}{c}}
   \begin{array}{l}
 x - 1 = t \\ 
 dx = dt \\ 
 \end{array} & \begin{array}{l}
 x + 1 = u \\ 
 dx = du \\ 
 \end{array}  \\
\end{array}} \right| = \frac{1}{2}\left( {\int {\frac{{dt}}{t}}  - \int {\frac{{du}}{u}} } \right) =  \\ 
  = \frac{1}{2}\left( {\ln t - \ln u} \right) = \frac{1}{2}\left( {\ln \left| {x - 1} \right| - \ln \left| {x + 1} \right|} \right) = \frac{1}{2}\ln \left| {\frac{{x - 1}}{{x + 1}}} \right| \\ 
 \end{array}\]