Question Number 220677 by fantastic last updated on 17/May/25
$$\int\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}\:}}{dx} \\ $$
Answered by Frix last updated on 17/May/25
![∫(√(x+(√(x^2 +1))))dx =^([t=(√(x+(√(x^2 +1))))]) =∫(t^2 +t^(−2) )dt=(t^3 /3)−t^(−1) = =((2(2x−(√(x^2 +1)))(√(x+(√(x^2 +1)))))/3)+C](https://www.tinkutara.com/question/Q220684.png)
$$\int\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dx}\:\overset{\left[{t}=\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\right]} {=} \\ $$$$=\int\left({t}^{\mathrm{2}} +{t}^{−\mathrm{2}} \right){dt}=\frac{{t}^{\mathrm{3}} }{\mathrm{3}}−{t}^{−\mathrm{1}} = \\ $$$$=\frac{\mathrm{2}\left(\mathrm{2}{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\right)\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}}{\mathrm{3}}+{C} \\ $$