∫ 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}} then d a d x = 2 x and d b d a = 1 2 a − 1 / 2 {\displaystyle {\frac {da}{dx}}=2x{\text{ and }}{\frac {db}{da}}={\frac {1}{2}}a^{-1/2}}
then d a d x d b d a = 2 x 1 2 a − 1 / 2 {\displaystyle {\frac {da}{dx}}{\frac {db}{da}}=2x{\frac {1}{2}}a^{-1/2}}
![{\displaystyle {\frac {d}{dx}}\left[(x^{2}+1)^{\frac {1}{2}}+c\right]}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/69232b433410babcdd65d4173dbeb99ca5d2a38d)
and 
then 
