∫ 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 ] = x x 2 + 1 {\displaystyle {\frac {d}{dx}}\left[(x^{2}+1)^{\frac {1}{2}}+c\right]={\frac {x}{\sqrt {x^{2}+1}}}}
a = x 2 + 1 x = 1 2 ( u + v ) v = 1 2 ( x − y ) y = 1 2 ( u − v ) {\displaystyle {\begin{aligned}a&=x^{2}+1&x&={\tfrac {1}{\sqrt {2}}}(u+v)\\[0.6ex]v&={\tfrac {1}{\sqrt {2}}}(x-y)\qquad &y&={\tfrac {1}{\sqrt {2}}}(u-v)\end{aligned}}}
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}}
d a d x ⋅ d b d a = ( 2 x ) ( 1 2 a − 1 / 2 ) = x a − 1 / 2 = x ( x 2 + 1 ) − 1 / 2 = x x 2 + 1 {\displaystyle {\frac {da}{dx}}\cdot {\frac {db}{da}}=\left(2x\right)\left({\frac {1}{2}}a^{-1/2}\right)=xa^{-1/2}=x(x^{2}+1)^{-1/2}={\frac {x}{\sqrt {x^{2}+1}}}}
![{\displaystyle {\frac {d}{dx}}\left[(x^{2}+1)^{\frac {1}{2}}+c\right]={\frac {x}{\sqrt {x^{2}+1}}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/3554e3a809bdd72518b30ec0c10b7cb78eba7739)
![{\displaystyle {\begin{aligned}a&=x^{2}+1&x&={\tfrac {1}{\sqrt {2}}}(u+v)\\[0.6ex]v&={\tfrac {1}{\sqrt {2}}}(x-y)\qquad &y&={\tfrac {1}{\sqrt {2}}}(u-v)\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/6c39c643f19670e6c8a4c6e8c0cfbed2231e3026)
and 
then 
