# x-3-x-4-dx-

Question Number 77290 by TawaTawa last updated on 05/Jan/20
$$\int\:\sqrt{\mathrm{x}^{\mathrm{3}} \:+\:\mathrm{x}^{\mathrm{4}} }\:\:\mathrm{dx} \\$$
Commented by mathmax by abdo last updated on 05/Jan/20
$${let}\:{I}\:=\int\sqrt{{x}^{\mathrm{3}} \:+{x}^{\mathrm{4}} }{dx}\:\Rightarrow\:{I}\:=\int\sqrt{{x}^{\mathrm{3}} \left(\mathrm{1}+{x}\right)}{dx}\:=\int\:{x}^{\frac{\mathrm{3}}{\mathrm{2}}} \sqrt{\mathrm{1}+{x}}{dx} \\$$$${changement}\:\sqrt{\mathrm{1}+{x}}={t}\:{hive}\:\mathrm{1}+{x}={t}^{\mathrm{2}} \:\Rightarrow \\$$$${I}\:=\int\left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {t}\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {dt}\:=_{{t}={ch}\left({u}\right)} \mathrm{2}\:\int{ch}^{\mathrm{2}} {u}\left({sh}^{\mathrm{2}} {u}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {sh}\left({u}\right){du} \\$$$$=\mathrm{2}\:\int\:{ch}^{\mathrm{2}} {u}\:{sh}^{\mathrm{4}} {u}\:{du}\:=\mathrm{2}\int\left(\frac{{sh}\left(\mathrm{2}{u}\right)}{\mathrm{2}}\right)^{\mathrm{2}} {sh}^{\mathrm{2}} {u}\:{du} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:{sh}^{\mathrm{2}} \left(\mathrm{2}{u}\right){sh}^{\mathrm{2}} \left({u}\right){du}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{ch}\left(\mathrm{4}{u}\right)−\mathrm{1}}{\mathrm{2}}×\frac{{ch}\left(\mathrm{2}{u}\right)−\mathrm{1}}{\mathrm{2}}{du} \\$$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\left({ch}\left(\mathrm{4}{u}\right)−\mathrm{1}\right)\left({ch}\left(\mathrm{2}{u}\right)−\mathrm{1}\right){du}\:\Rightarrow \\$$$$\mathrm{8}{I}\:=\int\left({ch}\left(\mathrm{4}{u}\right){ch}\left(\mathrm{2}{u}\right)−{ch}\left(\mathrm{4}{u}\right)−{ch}\left(\mathrm{2}{u}\right)+\mathrm{1}\right){du} \\$$$$=\int\:\:\frac{\left({e}^{\mathrm{4}{u}} +{e}^{−\mathrm{4}{u}} \right)\left({e}^{\mathrm{2}{u}} +{e}^{−\mathrm{2}{u}} \right)}{\mathrm{4}}{du}−\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{4}{u}\right)−\frac{\mathrm{1}}{\mathrm{2}}{sh}\left(\mathrm{2}{u}\right)\:+{u}\:+{c} \\$$$$=\frac{\mathrm{1}}{\mathrm{4}}\int\left({e}^{\mathrm{6}{u}} +{e}^{\mathrm{2}{u}} +{e}^{−\mathrm{2}{u}} \:+{e}^{−\mathrm{6}{u}} \right)−\frac{\mathrm{1}}{\mathrm{8}}\left({e}^{\mathrm{4}{u}} −{e}^{−\mathrm{4}{u}} \right)−\frac{\mathrm{1}}{\mathrm{2}}\left({e}^{\mathrm{2}{u}} −{e}^{−\mathrm{2}{u}} \right)\:+{u}\:+{c} \\$$$$=\frac{\mathrm{1}}{\mathrm{24}}{e}^{\mathrm{6}{u}} \:−\frac{\mathrm{1}}{\mathrm{24}}{e}^{−\mathrm{6}{u}} \:+\frac{\mathrm{1}}{\mathrm{8}}{e}^{\mathrm{2}{u}} −\frac{\mathrm{1}}{\mathrm{8}}{e}^{−\mathrm{2}{u}} −\frac{\mathrm{1}}{\mathrm{8}}{e}^{\mathrm{4}{u}} +\frac{\mathrm{1}}{\mathrm{8}}{e}^{−\mathrm{4}{u}} −\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{u}} \\$$$$+\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{u}} \:\:+{u}\:+{c} \\$$$$=\frac{\mathrm{1}}{\mathrm{24}}{e}^{\mathrm{6}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)} −\frac{\mathrm{1}}{\mathrm{24}}\:{e}^{−\mathrm{6}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)} \:+\frac{\mathrm{1}}{\mathrm{8}}{e}^{\mathrm{2}{ln}\left({g}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)} −\frac{\mathrm{1}}{\mathrm{8}}{e}^{−\mathrm{2}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)} \\$$$$−\frac{\mathrm{1}}{\mathrm{8}}{e}^{\mathrm{4}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)} \:+\frac{\mathrm{1}}{\mathrm{8}}{e}^{−\mathrm{4}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)} −\frac{\mathrm{1}}{\mathrm{2}}{e}^{\mathrm{2}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)} \:+\frac{\mathrm{1}}{\mathrm{2}}{e}^{−\mathrm{2}{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)} \\$$$$+{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)\:+{c} \\$$$$=\frac{\mathrm{1}}{\mathrm{24}}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{6}} \:−\frac{\mathrm{1}}{\mathrm{24}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{6}} }\:+\frac{\mathrm{1}}{\mathrm{8}}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:−\frac{\mathrm{1}}{\mathrm{8}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} } \\$$$$−\frac{\mathrm{1}}{\mathrm{8}}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{4}} \:+\frac{\mathrm{1}}{\mathrm{8}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{2}}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \:+\frac{\mathrm{1}}{\mathrm{2}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} } \\$$$$+{ln}\left({t}+\sqrt{{t}^{\mathrm{2}} −\mathrm{1}}\right)\:+{C} \\$$$$=\frac{\mathrm{1}}{\mathrm{24}}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}\right)^{\mathrm{6}} −\frac{\mathrm{1}}{\mathrm{24}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}\right)^{\mathrm{6}} }\:+\frac{\mathrm{1}}{\mathrm{8}}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}\right)^{\mathrm{2}} \\$$$$−\frac{\mathrm{1}}{\mathrm{8}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}\right)^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{8}}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}\right)^{\mathrm{4}} \:+\frac{\mathrm{1}}{\mathrm{8}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}\right)^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}\right)^{\mathrm{2}} \\$$$$+\frac{\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}\right)^{\mathrm{2}} }\:+{ln}\left(\sqrt{\mathrm{1}+{x}}+\sqrt{{x}}\right)\:+{C}\:. \\$$$$\\$$
Commented by TawaTawa last updated on 05/Jan/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir}. \\$$
Commented by mathmax by abdo last updated on 05/Jan/20
$${you}\:{are}\:{welcome}. \\$$
Answered by petrochengula last updated on 05/Jan/20
$$=\int{x}\sqrt{{x}+{x}^{\mathrm{2}} }{dx} \\$$$${x}+{x}^{\mathrm{2}} ={x}^{\mathrm{2}} +{x}+\frac{\mathrm{1}}{\mathrm{4}}−\frac{\mathrm{1}}{\mathrm{4}} \\$$$$=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}} \\$$$$=\int{x}\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}{dx} \\$$$$=\int{x}\sqrt{\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\frac{\mathrm{1}}{\mathrm{4}}}{dx}=\int{x}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}\left(\left(\mathrm{2}{x}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}\right)}{dx} \\$$$$=\frac{\mathrm{1}}{\mathrm{2}}\int{x}\sqrt{\left(\mathrm{2}{x}−\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}{dx} \\$$$${cosh}\:\theta=\mathrm{2}{x}−\mathrm{1}\Rightarrow\frac{{cosh}\theta+\mathrm{1}}{\mathrm{2}}={x} \\$$$$\frac{\mathrm{1}}{\mathrm{2}}{sinh}\theta{d}\theta={dx} \\$$$$=\frac{\mathrm{1}}{\mathrm{8}}\int\left({cosh}\theta+\mathrm{1}\right){sinh}^{\mathrm{2}} \theta{d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{8}}\int{sinh}^{\mathrm{2}} \theta{cosh}\theta{d}\theta+\frac{\mathrm{1}}{\mathrm{8}}\int{sinh}^{\mathrm{2}} \theta{d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{24}}{sinh}^{\mathrm{3}} \theta+\frac{\mathrm{1}}{\mathrm{8}}\int\frac{\mathrm{1}}{\mathrm{2}}\left({cosh}\mathrm{2}\theta−\mathrm{1}\right){d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{24}}{sinh}^{\mathrm{3}} \theta+\frac{\mathrm{1}}{\mathrm{16}}\int\left({cosh}\mathrm{2}\theta−\mathrm{1}\right){d}\theta \\$$$$=\frac{\mathrm{1}}{\mathrm{24}}{sinh}^{\mathrm{3}} \theta+\frac{\mathrm{1}}{\mathrm{32}}{sinh}\mathrm{2}\theta−\frac{\mathrm{1}}{\mathrm{16}}\theta+{C} \\$$$$=\frac{\mathrm{1}}{\mathrm{24}}\left({sinh}\left({cosh}^{−\mathrm{1}} \left(\mathrm{2}{x}−\mathrm{1}\right)\right)\right)^{\mathrm{3}} +\frac{\mathrm{1}}{\mathrm{32}}{sinh}\mathrm{2}\left({cosh}^{−\mathrm{1}} \left(\mathrm{2}{x}−\mathrm{1}\right)\right)−\frac{\mathrm{1}}{\mathrm{16}}{cosh}^{−\mathrm{1}} \left(\mathrm{2}{x}−\mathrm{1}\right)+{C} \\$$
Commented by petrochengula last updated on 05/Jan/20
$${please}\:{check} \\$$
Commented by TawaTawa last updated on 05/Jan/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\$$
Commented by petrochengula last updated on 05/Jan/20
$${thanks} \\$$
Answered by MJS last updated on 05/Jan/20
$$\int\sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{3}} }{dx}=\int{x}\sqrt{{x}\left({x}+\mathrm{1}\right)}{dx}= \\$$$$\:\:\:\:\:\left[{t}=\sqrt{{x}\left({x}+\mathrm{1}\right)}\:\rightarrow\:{dx}=\frac{\mathrm{2}\sqrt{{x}\left({x}+\mathrm{1}\right)}}{\mathrm{2}{x}+\mathrm{1}}{dt}\right] \\$$$$=\int{t}^{\mathrm{2}} {dt}−\int\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}{dt}= \\$$$$\\$$$$\:\:\:\:\:\int\frac{{t}^{\mathrm{2}} }{\:\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}{dt}= \\$$$$\:\:\:\:\:\:\:\:\:\:\left[{u}=\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{4}{t}}{du}\right] \\$$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8}}\int\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}{du}\:\mathrm{and}\:\mathrm{this}\:\mathrm{should}\:\mathrm{be}\:\mathrm{easy} \\$$$$\\$$$$=\frac{\mathrm{1}}{\mathrm{3}}{t}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{8}}{t}\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\left(\mathrm{2}{t}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} +\mathrm{1}}\right)\:= \\$$$$=\frac{\mathrm{1}}{\mathrm{24}}\left(\mathrm{8}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{3}\right)\sqrt{{x}\left({x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{16}}\mathrm{ln}\:\left(\mathrm{2}{x}+\mathrm{1}+\mathrm{2}\sqrt{{x}\left({x}+\mathrm{1}\right)}\right)\:+{C} \\$$
Commented by TawaTawa last updated on 05/Jan/20
$$\mathrm{God}\:\mathrm{bless}\:\mathrm{you}\:\mathrm{sir} \\$$