\[\begin{array}{l}
 T11. \to \int {\frac{{dx}}{{1 + {x^2}}}}  = {\mathop{\rm arctg}\nolimits} x + C =  - {\mathop{\rm arcctg}\nolimits} x + C \\ 
 {\rm{Dokaz}}\,{\rm{(Proof):}} \\ 
 \int {\frac{{dx}}{{1 + {x^2}}}}  = \left| \begin{array}{l}
 x = {\mathop{\rm tg}\nolimits} t = \frac{{\sin t}}{{\cos t}} \\ 
 dx = \frac{1}{{{{\cos }^2}t}}dt \\ 
 t = {\mathop{\rm arctg}\nolimits} x \\ 
 \end{array} \right| = \int {\frac{{\frac{1}{{{{\cos }^2}t}}}}{{1 + \frac{{{{\sin }^2}t}}{{{{\cos }^2}t}}}}dt}  = \int {\frac{{\frac{1}{{{{\cos }^2}t}}}}{{\frac{{{{\sin }^2}t + {{\cos }^2}t}}{{{{\cos }^2}t}}}}dt = \left| {{{\sin }^2}t + {{\cos }^2}t = 1} \right| = \int {dt}  = t = } {\mathop{\rm arctg}\nolimits} x \\ 
 \end{array}\]