∫ x x 2 + 1 d x = x 2 + 1 + c {\displaystyle \int {\frac {x}{\sqrt {x^{2}+1}}}dx={\sqrt {x^{2}+1}}+c}
d d x [ ( x 2 + 1 ) 1 2 + c ] {\displaystyle {\frac {d}{dx}}\left[(x^{2}+1)^{\frac {1}{2}}+c\right]}
let a = x 2 + 1 {\displaystyle a=x^{2}+1} and b = a 1 / 2 {\displaystyle b=a^{1/2}}