Question Number 77242 by Tony Lin last updated on 04/Jan/20
![prove that ∫_0 ^a (√(2+(a/x)−2(√(a/x)) ))dx=a[(1/(√2))ln((√2)+1)+1]](Q77242.png)
$${prove}\:{that} \\ $$$$\:\int_{\mathrm{0}} ^{{a}} \sqrt{\mathrm{2}+\frac{{a}}{{x}}−\mathrm{2}\sqrt{\frac{{a}}{{x}}}\:}{dx}={a}\left[\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}{ln}\left(\sqrt{\mathrm{2}}+\mathrm{1}\right)+\mathrm{1}\right] \\ $$
Commented by mathmax by abdo last updated on 04/Jan/20
![let f(a)=∫_0 ^a (√(2+(a/x)−2(√(a/x))))dx changement (√(a/x))=t hive (a/x) =t^2 ⇒a=xt^2 ⇒x=at^(−2) ⇒dx =−2at^(−3) dt and f(a)=∫_0 ^1 (√(2+t^2 −2t))(−2a)t^(−3 ) dt =(−2a)∫_0 ^1 (√(t^2 −2t+1+1))t^(−3) dt =−2a∫_0 ^1 (√((t−1)^2 +1))t^(−3) dt =_(t−1 =sh(u)) −2a ∫_(argsh(−1)) ^0 ch(u)(1+sh(u))^(−3) ch(u)du =−2a ∫_(−ln(1+(√2))) ^0 ch^2 u(1+sh(u))^2 du =−2a ∫_(−ln(1+(√2))) ^0 ch^2 u(1+2shu +sh^2 u)du =−2a∫_(−ln(1+(√2))) ^0 ch^2 u du −4a ∫_(−ln(1+(√2))) ^0 shu ch^2 udu −2a∫_(−ln(1+(√2))) ^0 sh^2 u ch^2 u du =−2a∫_(−ln(1+(√2))) ^0 ((ch(2u)+1)/2)du −4a [(1/3)ch^3 u]_(−ln(1+(√2))) ^0 −(a/2) ∫_(−ln(1+(√2))) ^0 sh^2 u du =a [u+(1/2)shu]_(−ln(1+(√2))) ^0 −((4a)/3)[((e^(3u) +e^(−3u) )/2)]_(−ln(1+(√2))) ^0 −(a/2)∫_(−ln(1+(√2))) ^0 ((ch(2u)−1)/2)du rest to achieve the calculus...](Q77262.png)
$${let}\:{f}\left({a}\right)=\int_{\mathrm{0}} ^{{a}} \sqrt{\mathrm{2}+\frac{{a}}{{x}}−\mathrm{2}\sqrt{\frac{{a}}{{x}}}}{dx}\:\:{changement}\:\sqrt{\frac{{a}}{{x}}}={t}\:{hive} \\ $$$$\frac{{a}}{{x}}\:={t}^{\mathrm{2}} \:\Rightarrow{a}={xt}^{\mathrm{2}} \:\Rightarrow{x}={at}^{−\mathrm{2}} \:\Rightarrow{dx}\:=−\mathrm{2}{at}^{−\mathrm{3}} \:{dt}\:{and} \\ $$$${f}\left({a}\right)=\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\mathrm{2}+{t}^{\mathrm{2}} −\mathrm{2}{t}}\left(−\mathrm{2}{a}\right){t}^{−\mathrm{3}\:} {dt} \\ $$$$=\left(−\mathrm{2}{a}\right)\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{{t}^{\mathrm{2}} −\mathrm{2}{t}+\mathrm{1}+\mathrm{1}}{t}^{−\mathrm{3}} \:{dt}\:=−\mathrm{2}{a}\int_{\mathrm{0}} ^{\mathrm{1}} \sqrt{\left({t}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{1}}{t}^{−\mathrm{3}} \:{dt} \\ $$$$=_{{t}−\mathrm{1}\:={sh}\left({u}\right)} \:\:−\mathrm{2}{a}\:\int_{{argsh}\left(−\mathrm{1}\right)} ^{\mathrm{0}} {ch}\left({u}\right)\left(\mathrm{1}+{sh}\left({u}\right)\right)^{−\mathrm{3}} \:{ch}\left({u}\right){du} \\ $$$$=−\mathrm{2}{a}\:\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{ch}^{\mathrm{2}} {u}\left(\mathrm{1}+{sh}\left({u}\right)\right)^{\mathrm{2}} \:{du} \\ $$$$=−\mathrm{2}{a}\:\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{ch}^{\mathrm{2}} {u}\left(\mathrm{1}+\mathrm{2}{shu}\:+{sh}^{\mathrm{2}} {u}\right){du} \\ $$$$=−\mathrm{2}{a}\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{ch}^{\mathrm{2}} {u}\:{du}\:−\mathrm{4}{a}\:\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{shu}\:{ch}^{\mathrm{2}} {udu} \\ $$$$−\mathrm{2}{a}\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{sh}^{\mathrm{2}} {u}\:{ch}^{\mathrm{2}} {u}\:{du} \\ $$$$=−\mathrm{2}{a}\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \frac{{ch}\left(\mathrm{2}{u}\right)+\mathrm{1}}{\mathrm{2}}{du}\:−\mathrm{4}{a}\:\left[\frac{\mathrm{1}}{\mathrm{3}}{ch}^{\mathrm{3}} {u}\right]_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \\ $$$$−\frac{{a}}{\mathrm{2}}\:\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:{sh}^{\mathrm{2}} {u}\:{du}\: \\ $$$$={a}\:\left[{u}+\frac{\mathrm{1}}{\mathrm{2}}{shu}\right]_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} −\frac{\mathrm{4}{a}}{\mathrm{3}}\left[\frac{{e}^{\mathrm{3}{u}} +{e}^{−\mathrm{3}{u}} }{\mathrm{2}}\right]_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \\ $$$$−\frac{{a}}{\mathrm{2}}\int_{−{ln}\left(\mathrm{1}+\sqrt{\mathrm{2}}\right)} ^{\mathrm{0}} \:\frac{{ch}\left(\mathrm{2}{u}\right)−\mathrm{1}}{\mathrm{2}}{du}\:\:{rest}\:{to}\:{achieve}\:{the}\:{calculus}... \\ $$
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≥0∧x>0 ∨ a≤0∧x<0 ⇒ ∫(√(2+(a/x)−2(√(a/x))))dx=∫((√(2x−2(√(ax))+a))/(√x))dx= [t=(√x) → dx=2(√x)dt] =2∫(√(t^2 −2(√a)t+a))dt= =((2t−(√a))/2)(√(2t^2 −2(√a)t+a))+((√2)/4)aln (2t−(√a)+(√(4t^2 −4(√a)t+2a))) = =((2(√x)−(√a))/2)(√(2x−2(√(ax))+a))+((√2)/4)aln (2(√x)−(√a)+(√(4x−4(√(ax))+2a))) +C now just insert the values x=[0; a]](Q77265.png)
$${a}\geqslant\mathrm{0}\wedge{x}>\mathrm{0}\:\vee\:{a}\leqslant\mathrm{0}\wedge{x}<\mathrm{0} \\ $$$$\Rightarrow \\ $$$$\int\sqrt{\mathrm{2}+\frac{{a}}{{x}}−\mathrm{2}\sqrt{\frac{{a}}{{x}}}}{dx}=\int\frac{\sqrt{\mathrm{2}{x}−\mathrm{2}\sqrt{{ax}}+{a}}}{\sqrt{{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\ $$$$=\mathrm{2}\int\sqrt{{t}^{\mathrm{2}} −\mathrm{2}\sqrt{{a}}{t}+{a}}{dt}= \\ $$$$=\frac{\mathrm{2}{t}−\sqrt{{a}}}{\mathrm{2}}\sqrt{\mathrm{2}{t}^{\mathrm{2}} −\mathrm{2}\sqrt{{a}}{t}+{a}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}{a}\mathrm{ln}\:\left(\mathrm{2}{t}−\sqrt{{a}}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} −\mathrm{4}\sqrt{{a}}{t}+\mathrm{2}{a}}\right)\:= \\ $$$$=\frac{\mathrm{2}\sqrt{{x}}−\sqrt{{a}}}{\mathrm{2}}\sqrt{\mathrm{2}{x}−\mathrm{2}\sqrt{{ax}}+{a}}+\frac{\sqrt{\mathrm{2}}}{\mathrm{4}}{a}\mathrm{ln}\:\left(\mathrm{2}\sqrt{{x}}−\sqrt{{a}}+\sqrt{\mathrm{4}{x}−\mathrm{4}\sqrt{{ax}}+\mathrm{2}{a}}\right)\:+{C} \\ $$$$\mathrm{now}\:\mathrm{just}\:\mathrm{insert}\:\mathrm{the}\:\mathrm{values}\:{x}=\left[\mathrm{0};\:{a}\right] \\ $$
Commented by Tony Lin last updated on 05/Jan/20
$${thanks}\:{sir} \\ $$