# prove-that-0-a-2-a-x-2-a-x-dx-a-1-2-ln-2-1-1-

Question Number 77242 by Tony Lin last updated on 04/Jan/20
$${prove}\:{that} \\$$$$\:\int_{\mathrm{0}} ^{{a}} \sqrt{\mathrm{2}+\frac{{a}}{{x}}−\mathrm{2}\sqrt{\frac{{a}}{{x}}}\:}{dx}={a}\left[\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+\mathrm{1}\right] \\$$
Commented by mathmax by abdo last updated on 04/Jan/20
$${let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{{a}} \sqrt{\mathrm{2}+\frac{{a}}{{x}}−\mathrm{2}\sqrt{\frac{{a}}{{x}}}}{dx}\:\:{changement}\:\sqrt{\frac{{a}}{{x}}}={t}\:{hive} \\$$$$\frac{{a}}{{x}}\:={t}^{\mathrm{2}} \:\Rightarrow{a}={xt}^{\mathrm{2}} \:\Rightarrow{x}={at}^{−\mathrm{2}} \:\Rightarrow{dx}\:=−\mathrm{2}{at}^{−\mathrm{3}} \:{dt}\:{and} \\$$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{2}+{t}^{\mathrm{2}} −\mathrm{2}{t}}\left(−\mathrm{2}{a}\right){t}^{−\mathrm{3}\:} {dt} \\$$$$=\left(−\mathrm{2}{a}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}+\mathrm{1}}{t}^{−\mathrm{3}} \:{dt}\:=−\mathrm{2}{a}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left({t}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}{t}^{−\mathrm{3}} \:{dt} \\$$$$=_{{t}−\mathrm{1}\:={sh}\left({u}\right)} \:\:−\mathrm{2}{a}\:\int_{{argsh}\left(−\mathrm{1}\right)} ^{\mathrm{0}} {ch}\left({u}\right)\left(\mathrm{1}+{sh}\left({u}\right)\right)^{−\mathrm{3}} \:{ch}\left({u}\right){du} \\$$$$=−\mathrm{2}{a}\:\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{ch}^{\mathrm{2}} {u}\left(\mathrm{1}+{sh}\left({u}\right)\right)^{\mathrm{2}} \:{du} \\$$$$=−\mathrm{2}{a}\:\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{ch}^{\mathrm{2}} {u}\left(\mathrm{1}+\mathrm{2}{shu}\:+{sh}^{\mathrm{2}} {u}\right){du} \\$$$$=−\mathrm{2}{a}\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{ch}^{\mathrm{2}} {u}\:{du}\:−\mathrm{4}{a}\:\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{shu}\:{ch}^{\mathrm{2}} {udu} \\$$$$−\mathrm{2}{a}\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{sh}^{\mathrm{2}} {u}\:{ch}^{\mathrm{2}} {u}\:{du} \\$$$$=−\mathrm{2}{a}\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \frac{{ch}\left(\mathrm{2}{u}\right)+\mathrm{1}}{\mathrm{2}}{du}\:−\mathrm{4}{a}\:\left[\frac{\mathrm{1}}{\mathrm{3}}{ch}^{\mathrm{3}} {u}\right]_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \\$$$$−\frac{{a}}{\mathrm{2}}\:\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{sh}^{\mathrm{2}} {u}\:{du}\: \\$$$$={a}\:\left[{u}+\frac{\mathrm{1}}{\mathrm{2}}{shu}\right]_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} −\frac{\mathrm{4}{a}}{\mathrm{3}}\left[\frac{{e}^{\mathrm{3}{u}} +{e}^{−\mathrm{3}{u}} }{\mathrm{2}}\right]_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \\$$$$−\frac{{a}}{\mathrm{2}}\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:\frac{{ch}\left(\mathrm{2}{u}\right)−\mathrm{1}}{\mathrm{2}}{du}\:\:{rest}\:{to}\:{achieve}\:{the}\:{calculus}… \\$$
Commented by Tony Lin last updated on 05/Jan/20
$${thanks}\:{sir} \\$$
Commented by mathmax by abdo last updated on 05/Jan/20
$${you}\:{are}\:{welcome}. \\$$
Answered by MJS last updated on 05/Jan/20
$${a}\geqslant\mathrm{0}\wedge{x}>\mathrm{0}\:\vee\:{a}\leqslant\mathrm{0}\wedge{x}<\mathrm{0} \\$$$$\Rightarrow \\$$$$\int\sqrt{\mathrm{2}+\frac{{a}}{{x}}−\mathrm{2}\sqrt{\frac{{a}}{{x}}}}{dx}=\int\frac{\sqrt{\mathrm{2}{x}−\mathrm{2}\sqrt{{ax}}+{a}}}{\:\sqrt{{x}}}{dx}= \\$$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\$$$$=\mathrm{2}\int\sqrt{{t}^{\mathrm{2}} −\mathrm{2}\sqrt{{a}}{t}+{a}}{dt}= \\$$$$=\frac{\mathrm{2}{t}−\sqrt{{a}}}{\mathrm{2}}\sqrt{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}\sqrt{{a}}{t}+{a}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}{a}\mathrm{ln}\:\left(\mathrm{2}{t}−\sqrt{{a}}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} −\mathrm{4}\sqrt{{a}}{t}+\mathrm{2}{a}}\right)\:= \\$$$$=\frac{\mathrm{2}\sqrt{{x}}−\sqrt{{a}}}{\mathrm{2}}\sqrt{\mathrm{2}{x}−\mathrm{2}\sqrt{{ax}}+{a}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}{a}\mathrm{ln}\:\left(\mathrm{2}\sqrt{{x}}−\sqrt{{a}}+\sqrt{\mathrm{4}{x}−\mathrm{4}\sqrt{{ax}}+\mathrm{2}{a}}\right)\:+{C} \\$$$$\mathrm{now}\:\mathrm{just}\:\mathrm{insert}\:\mathrm{the}\:\mathrm{values}\:{x}=\left[\mathrm{0};\:{a}\right] \\$$
Commented by Tony Lin last updated on 05/Jan/20
$${thanks}\:{sir} \\$$