5.4 Indefinite Integrals and the Net Change Theorem/1

$\int {\frac {x}{\sqrt {x^{2}+1}}}dx={\sqrt {x^{2}+1}}+c$

Show that: ${\frac {d}{dx}}\left[(x^{2}+1)^{\frac {1}{2}}+c\right]={\frac {x}{\sqrt {x^{2}+1}}}$

${\begin{aligned}a&=x^{2}+1&b&=a^{1/2}\\[0.6ex]{\frac {da}{dx}}&=2xy&{\frac {db}{da}}&={\frac {1}{2}}a^{-1/2}\end{aligned}}$

${\frac {da}{dx}}\cdot {\frac {db}{da}}=\left(2x\right)\left({\frac {1}{2}}a^{-1/2}\right)=xa^{-1/2}=x(x^{2}+1)^{-1/2}={\frac {x}{\sqrt {x^{2}+1}}}$

$\int {\frac {x}{\sqrt {x^{2}+1}}}dx={\sqrt {x^{2}+1}}+c$

${\frac {d}{dx}}\left[(x^{2}+1)^{\frac {1}{2}}+c\right]={\frac {x}{\sqrt {x^{2}+1}}}$

${\begin{aligned}a&=x^{2}+1&b&=a^{1/2}\\[0.6ex]{\frac {da}{dx}}&=2xy&{\frac {db}{da}}&={\frac {1}{2}}a^{-1/2}\end{aligned}}$

${\frac {da}{dx}}\cdot {\frac {db}{da}}=\left(2x\right)\left({\frac {1}{2}}a^{-1/2}\right)=xa^{-1/2}=x(x^{2}+1)^{-1/2}={\frac {x}{\sqrt {x^{2}+1}}}$