Question Number 25229 by NECx last updated on 06/Dec/17
$$\int\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dx} \\ $$
Commented by prakash jain last updated on 06/Dec/17
$${x}=\mathrm{sinh}\:{u} \\ $$$${dx}=\mathrm{cosh}\:{u}\centerdot{du} \\ $$$$\mathrm{1}+\mathrm{sinh}^{\mathrm{2}} {u}=\mathrm{cosh}^{\mathrm{2}} {u} \\ $$$$\int\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dx}=\int\frac{\mathrm{cosh}\:{u}}{\mathrm{cosh}\:{u}}{du} \\ $$$$={u} \\ $$$$=\mathrm{sinh}^{−\mathrm{1}} {x} \\ $$$$=\mathrm{ln}\:\left(\mathrm{1}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$