Question Number 85600 by M±th+et£s last updated on 23/Mar/20
$$\int\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:{dx} \\ $$
Commented by mathmax by abdo last updated on 23/Mar/20
$${I}\:=\int\:\frac{{x}^{\mathrm{2}} }{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}\:=\int\:\:\frac{{x}^{\mathrm{2}} +\mathrm{1}−\mathrm{1}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{dx}\:=\int\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}\:\:−\int\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }} \\ $$$$\int\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }{dx}\:=_{{x}={sh}\left({t}\right)} \:\:\:\int{ch}\left({t}\right){ch}\left({t}\right){dt}\:=\int\:{ch}^{\mathrm{2}} \left({t}\right){dt}=\int\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt} \\ $$$$=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:+{c}\:=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({t}\right){ch}\left({t}\right)\:=\frac{\mathrm{1}}{\mathrm{2}}{argsh}\left({x}\right)+\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+{c} \\ $$$$\int\:\:\frac{{dx}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\:=_{{x}={sh}\left({t}\right)} \:\:\:\int\:\:\frac{{ch}\left({t}\right)}{{ch}\left({t}\right)}{dt}\:={t}\:+\lambda\:\:={argsh}\left({x}\right)+\lambda\:= \\ $$$$={ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:+\lambda\:\Rightarrow{I}\:=−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\frac{{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}{\mathrm{2}}\:+{C} \\ $$
Commented by M±th+et£s last updated on 23/Mar/20
$${thank}\:{you}\:{sir} \\ $$
Commented by abdomathmax last updated on 23/Mar/20
$${you}\:{are}\:{welcome} \\ $$
Answered by john santu last updated on 23/Mar/20
![∫(1/2)x (((2x)/( (√(1+x^2 ))))) dx = ∫ (1/2)x (((d(1+x^2 ))/( (√(1+x^2 ))))) = (1/2)x (2(√(1+x^2 ))) − ∫ (√(1+x^2 )) dx= x(√(1+x^2 )) − ∫ (√(1+tan^2 u)) sec^2 x dx x(√(1+x^2 )) − ∫ sec^3 x dx = [ x=tan u ] x(√(1+x^2 )) − ((1/2)sec utan u+(1/2)ln∣sec u+tan u∣)+c (1/2)x(√(1+x^2 )) − (1/2)ln∣ x+(√(1+x^2 )) ∣ +c](https://www.tinkutara.com/question/Q85604.png)
$$\int\frac{\mathrm{1}}{\mathrm{2}}{x}\:\left(\frac{\mathrm{2}{x}}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\:{dx}\:=\: \\ $$$$\int\:\frac{\mathrm{1}}{\mathrm{2}}{x}\:\left(\frac{{d}\left(\mathrm{1}+{x}^{\mathrm{2}} \right)}{\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }}\right)\:= \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}\:\left(\mathrm{2}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)\:−\:\int\:\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:{dx}= \\ $$$${x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:−\:\int\:\sqrt{\mathrm{1}+\mathrm{tan}\:^{\mathrm{2}} {u}}\:\mathrm{sec}\:^{\mathrm{2}} {x}\:{dx} \\ $$$${x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:−\:\int\:\mathrm{sec}\:^{\mathrm{3}} {x}\:{dx}\:=\:\left[\:{x}=\mathrm{tan}\:{u}\:\right] \\ $$$${x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:−\:\left(\frac{\mathrm{1}}{\mathrm{2}}\mathrm{sec}\:{u}\mathrm{tan}\:{u}+\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\mathrm{sec}\:{u}+\mathrm{tan}\:{u}\mid\right)+{c} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}{x}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:−\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\mid\:{x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:\mid\:+{c} \\ $$
Commented by mathmax by abdo last updated on 23/Mar/20
$${where}\:{are}\:{you}\:{from}\:{sir}\:{john}… \\ $$
Commented by M±th+et£s last updated on 23/Mar/20
$${thsnm}\:{you}\:{sir} \\ $$