Question Number 82510 by jagoll last updated on 22/Feb/20
$$\int\:\sqrt{{x}+\sqrt{{x}}\:}\:{dx}\:=\:? \\ $$
Commented by mathmax by abdo last updated on 23/Feb/20
$${I}\:=\int\sqrt{{x}+\sqrt{{x}}}{dx}\:{changement}\:\sqrt{{x}}={t}\:{give}\:{x}={t}^{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\int\sqrt{{t}^{\mathrm{2}} +{t}}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\int\:{t}\sqrt{{t}^{\mathrm{2}} +{t}}{dt}\:=\mathrm{2}\int{t}\sqrt{{t}^{\mathrm{2}} +\mathrm{2}\frac{\mathrm{1}}{\mathrm{2}}{t}\:+\frac{\mathrm{1}}{\mathrm{4}\:}−\frac{\mathrm{1}}{\mathrm{4}}}{dt} \\ $$$$=\mathrm{2}\int\:{t}\sqrt{\left({t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}{dt}\:\:=_{{t}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\mathrm{1}}{\mathrm{2}}{ch}\left({u}\right)} \mathrm{2}\int\:\frac{\mathrm{1}}{\mathrm{2}}\left({ch}\left({u}\right)−\mathrm{1}\right)×\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({u}\right)\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\:\int\:\left({ch}\left({u}\right)−\mathrm{1}\right){sh}^{\mathrm{2}} \left({u}\right){du} \\ $$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\:{ch}\left({u}\right){sh}^{\mathrm{2}} \left({u}\right){du}\:−\frac{\mathrm{1}}{\mathrm{4}}\int\:\frac{{ch}\left(\mathrm{2}{u}\right)−\mathrm{1}}{\mathrm{2}}{du} \\ $$$$\left.=\frac{\mathrm{1}}{\mathrm{12}}{sh}^{\mathrm{3}} \left({u}\right)−\frac{\mathrm{1}}{\mathrm{8}}\:\int\:{ch}\left(\mathrm{2}{u}\right)+\frac{\mathrm{1}}{\mathrm{8}}{u}\:\right)+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}{sh}^{\mathrm{2}} {u}\:{shu}\:−\frac{\mathrm{1}}{\mathrm{16}}{sh}\left(\mathrm{2}{u}\right)+\frac{{u}}{\mathrm{8}}\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right)\sqrt{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{8}}{argch}\left(\mathrm{2}{t}+\mathrm{1}\right)\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}\right)\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}}−\frac{\mathrm{1}}{\mathrm{8}}\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{4}{t}}+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{2}{t}+\mathrm{1}+\sqrt{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}\right)\:+{C} \\ $$$$\left.=\frac{\mathrm{2}}{\mathrm{3}}\left({t}^{\mathrm{2}} \:+{t}\right)\sqrt{{t}^{\mathrm{2}} +{t}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} \:+{t}}\:+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{2}{t}+\mathrm{1}\right)+\mathrm{2}\sqrt{{t}^{\mathrm{2}} \:+{t}}\right)+{C} \\ $$$$\left.\Rightarrow{I}\:=\frac{\mathrm{2}}{\mathrm{3}}\left({x}+\sqrt{{x}}\right)\sqrt{{x}+\sqrt{{x}}}−\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{2}\sqrt{{x}}+\mathrm{1}\right)\sqrt{{x}+\sqrt{{x}}}+\frac{\mathrm{1}}{\mathrm{8}}{ln}\left(\mathrm{2}\sqrt{{x}}+\mathrm{1}+\mathrm{2}\sqrt{{x}+\sqrt{{x}}}\right)\:\right)\:+{C} \\ $$
Answered by john santu last updated on 22/Feb/20