\[\begin{array}{l}
 T16. \to \int {\frac{{dx}}{{1 - {x^2}}}}  = \frac{1}{2}\ln \left| {\frac{{1 + x}}{{1 - x}}} \right| + C = {\mathop{\rm arctgh}\nolimits} x + C \\ 
 {\rm{Dokaz}}\,{\rm{(Proof):}} \\ 
 \int {\frac{{dx}}{{1 - {x^2}}}}  = \int {\frac{1}{{(1 - x)(1 + x)}}dx}  = \frac{1}{2}\int {\frac{{2 + x - x}}{{(1 - x)(1 + x)}}dx}  = \frac{1}{2}\int {\frac{{(1 + x) + (1 - x)}}{{(1 - x)(1 + x)}}dx = }  \\ 
  = \frac{1}{2}\int {\frac{{(1 + x)}}{{(1 - x)(1 + x)}}dx}  + \frac{1}{2}\int {\frac{{(1 - x)}}{{(1 - x)(1 + x)}}dx = } \frac{1}{2}\left( {\int {\frac{1}{{1 - x}}dx}  + \int {\frac{1}{{1 + x}}dx} } \right) = \left| {\begin{array}{*{20}{c}}
   \begin{array}{l}
 1 - x = t \\ 
  - dx = dt \\ 
 \end{array} & \begin{array}{l}
 1 + x = u \\ 
 dx = du \\ 
 \end{array}  \\
\end{array}} \right| =  \\ 
  = \frac{1}{2}\left( {\int {\frac{1}{t}( - dt)}  + \int {\frac{1}{u}du} } \right) = \frac{1}{2}\left( { - \ln t + \ln u} \right) = \frac{1}{2}\left( { - \ln \left| {1 - x} \right| + \ln \left| {1 + x} \right|} \right) = \frac{1}{2}\left( {\ln \left| {1 + x} \right| - \ln \left| {1 - x} \right|} \right) =  \\ 
  = \frac{1}{2}\ln \left| {\frac{{1 + x}}{{1 - x}}} \right| \\ 
 \end{array}\]