Question Number 166301 by LEKOUMA last updated on 18/Feb/22
$$\int\frac{{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} +\sqrt{\left(\mathrm{1}+{x}^{\mathrm{2}} \right)^{\mathrm{3}} }}}{dx} \\ $$
Answered by MJS_new last updated on 18/Feb/22
![∫(x/( (√(x^2 +1+(x^2 +1)^(3/2) ))))dx= [t=x+(√(x^2 +1)) → dx=((√(x^2 +1))/(x+(√(x^2 +1))))dt] =((√2)/2)∫((t−1)/t^(3/2) )dt=((√2)/2)(2t^(1/2) +(2/t^(1/2) ))=(((√2)(t+1))/( (√t)))= ... =2(√(1+(√(x^2 +1))))+C](https://www.tinkutara.com/question/Q166303.png)
$$\int\frac{{x}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}+\left({x}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} }}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dx}=\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{dt}\right] \\ $$$$=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\frac{{t}−\mathrm{1}}{{t}^{\mathrm{3}/\mathrm{2}} }{dt}=\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\mathrm{2}{t}^{\mathrm{1}/\mathrm{2}} +\frac{\mathrm{2}}{{t}^{\mathrm{1}/\mathrm{2}} }\right)=\frac{\sqrt{\mathrm{2}}\left({t}+\mathrm{1}\right)}{\:\sqrt{{t}}}= \\ $$$$… \\ $$$$=\mathrm{2}\sqrt{\mathrm{1}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}+{C} \\ $$
Commented by cortano1 last updated on 18/Feb/22
$$\mathrm{Euler}\:\mathrm{subtitution} \\ $$
Commented by peter frank last updated on 19/Feb/22
$$\mathrm{thank}\:\mathrm{you} \\ $$
Answered by cortano1 last updated on 18/Feb/22
$$\:\int\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} +\sqrt{\left(\mathrm{1}+\mathrm{x}^{\mathrm{2}} \right)^{\mathrm{3}} }}}\:\left(\mathrm{x}\:\mathrm{dx}\right) \\ $$$$\:\mathrm{let}\:\mathrm{y}=\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }\:\Rightarrow\mathrm{y}^{\mathrm{2}} =\mathrm{1}+\mathrm{x}^{\mathrm{2}} \\ $$$$\Rightarrow\mathrm{y}\:\mathrm{dy}\:=\:\mathrm{x}\:\mathrm{dx} \\ $$$$\mathrm{I}=\int\:\frac{\mathrm{y}\:\mathrm{dy}}{\:\sqrt{\mathrm{y}^{\mathrm{2}} +\mathrm{y}^{\mathrm{3}} }}\:=\:\int\:\frac{\mathrm{dy}}{\:\sqrt{\mathrm{1}+\mathrm{y}}} \\ $$$$\mathrm{I}=\int\:\left(\mathrm{1}+\mathrm{y}\right)^{−\frac{\mathrm{1}}{\mathrm{2}}} \mathrm{dy}\:=\:\mathrm{2}\sqrt{\mathrm{1}+\mathrm{y}}\:+\:\mathrm{c} \\ $$$$\mathrm{I}=\mathrm{2}\sqrt{\mathrm{1}+\sqrt{\mathrm{1}+\mathrm{x}^{\mathrm{2}} }}\:+\:\mathrm{c}\: \\ $$
Commented by LEKOUMA last updated on 18/Feb/22
$${Thanks} \\ $$