\[\begin{array}{l}
 T18. \to \int {\frac{{dx}}{{\sqrt {{x^2} + 1} }}}  = \ln \left( {x + \sqrt {{x^2} + 1} } \right) + C = {\mathop{\rm arcsinh}\nolimits} x + C \\ 
 {\rm{Dokaz}}\,{\rm{(Proof):}} \\ 
 \int {\frac{{dx}}{{\sqrt {{x^2} + 1} }}}  = \left| \begin{array}{l}
 x = \sinh t \\ 
 dx = \cosh t \\ 
 {\mathop{\rm arcsinh}\nolimits} x = t \\ 
 \end{array} \right| = \int {\frac{{\cosh t}}{{\sqrt {{{\sinh }^2}t + 1} }}dx}  = \left| {{{\cosh }^2}t - {{\sinh }^2}t = 1} \right| = \int {\frac{{\cosh t}}{{\cosh t}}dx}  = \int {dt}  = t = {\mathop{\rm arcsinh}\nolimits} x \\ 
 \sinh x = \frac{{{e^x} - {e^{ - x}}}}{2} \to \left| {x = \sinh t} \right| \to x = \frac{{{e^t} - {e^{ - t}}}}{2} \\ 
 2x = {e^t} - {e^{ - t}}/ \cdot {e^t} \\ 
 2x{e^t} = {e^{2t}} - 1 \\ 
 {({e^t})^2} - 2x{e^t} - 1 = 0 \to \left| \begin{array}{l}
 a{x^2} + bx + c = 0 \\ 
 {x_{1/2}} = \frac{{ - b \pm \sqrt {{b^2} - 4ac} }}{{2a}} \\ 
 \end{array} \right| \to e_{1/2}^t = \frac{{2x \pm \sqrt {4{x^2} + 4} }}{2} = x \pm \sqrt {{x^2} + 1}  \\ 
 {e^t} = x + \sqrt {{x^2} + 1} \,\,\,/\ln  \to \ln {e^t} = \ln (x + \sqrt {{x^2} + 1} ) \\ 
 t = \ln (x + \sqrt {{x^2} + 1} ) \to \int {\frac{{dx}}{{\sqrt {{x^2} + 1} }}}  = t = \ln \left( {x + \sqrt {{x^2} + 1} } \right) \\ 
 \end{array}\]