Menu Close

M-x-x-2-5-dx-




Question Number 129212 by bramlexs22 last updated on 13/Jan/21
 M = ∫ (√(x+(√(x^2 +5)))) dx ?
$$\:\mathrm{M}\:=\:\int\:\sqrt{\mathrm{x}+\sqrt{\mathrm{x}^{\mathrm{2}} +\mathrm{5}}}\:\mathrm{dx}\:? \\ $$
Answered by TheSupreme last updated on 13/Jan/21
x+(√(x^2 +5))=t
$${x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}={t} \\ $$
Answered by MJS_new last updated on 13/Jan/21
∫(√(x+(√(x^2 +5)))) dx=       [t=(√(x+(√(x^2 +5)))) → dx=((t^4 +5)/t^3 )dt]  =∫((t^4 +5)/t^2 )dt=((t^4 −15)/(3t))=  =((2(2x−(√(x^2 +5)))(√(x+(√(x^2 +5)))))/3)+C
$$\int\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}\:{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}\:\rightarrow\:{dx}=\frac{{t}^{\mathrm{4}} +\mathrm{5}}{{t}^{\mathrm{3}} }{dt}\right] \\ $$$$=\int\frac{{t}^{\mathrm{4}} +\mathrm{5}}{{t}^{\mathrm{2}} }{dt}=\frac{{t}^{\mathrm{4}} −\mathrm{15}}{\mathrm{3}{t}}= \\ $$$$=\frac{\mathrm{2}\left(\mathrm{2}{x}−\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}\right)\sqrt{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{5}}}}{\mathrm{3}}+{C} \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *