∫ x x 2 + 1 d x = x 2 + 1 + c {\displaystyle \int {\frac {x}{\sqrt {x^{2}+1}}}dx={\sqrt {x^{2}+1}}+c}
Show that: 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}}}}
Let: a = x 2 + 1 b = a 1 / 2 d a d x = 2 x y d b d a = 1 2 a − 1 / 2 {\displaystyle {\begin{aligned}a&=x^{2}+1&b&=a^{1/2}\\[0.6ex]{\frac {da}{dx}}&=2xy&{\frac {db}{da}}&={\frac {1}{2}}a^{-1/2}\end{aligned}}}
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&b&=a^{1/2}\\[0.6ex]{\frac {da}{dx}}&=2xy&{\frac {db}{da}}&={\frac {1}{2}}a^{-1/2}\end{aligned}}}](https://en.wikipedia.org/api/rest_v1/media/math/render/svg/a4c2a8225c7919960fcaa97d8de3dcf1fc469549)
