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Evaluate-x-x-1-dx-




Question Number 67471 by AnjanDey last updated on 27/Aug/19
Evaluate:∫(√(x(√(x+1)))) dx
$$\boldsymbol{{Evaluate}}:\int\sqrt{{x}\sqrt{{x}+\mathrm{1}}}\:{dx} \\ $$
Commented by MJS last updated on 28/Aug/19
∫(√(x(√(x+1))))dx=       [t=(√x) → dx=2(√x)dt]  =2∫t^2 (t^2 +1)^(1/4) dt=  =−(4/(21))∫(dt/((t^2 +1)^(3/4) ))+2∫((t^4 +t^2 +(2/(21)))/((t^2 +1)^(3/4) ))dt  =−(4/(21))∫(dt/((t^2 +1)^(3/4) ))+(4/(21))t(3t^2 +1)(t^2 +1)^(1/4)   ...I cannot solve the 1^(st)  one  plus I somehow solved the 2^(nd)  one but I really  don′t remember how... I lost some papers...  anyway (d/dt)[(4/(21))t(3t^2 +1)(t^2 +1)^(1/4) ]=2((t^4 +t^2 +(2/(21)))/((t^2 +1)^(3/4) ))  and −(4/(21(t^2 +1)^(3/4) ))+2((t^4 +t^2 +(2/(21)))/((t^2 +1)^(3/4) ))=2t^2 (t^2 +1)^(1/4)
$$\int\sqrt{{x}\sqrt{{x}+\mathrm{1}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\ $$$$=\mathrm{2}\int{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} {dt}= \\ $$$$=−\frac{\mathrm{4}}{\mathrm{21}}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} }+\mathrm{2}\int\frac{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{21}}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} }{dt} \\ $$$$=−\frac{\mathrm{4}}{\mathrm{21}}\int\frac{{dt}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} }+\frac{\mathrm{4}}{\mathrm{21}}{t}\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$$$…\mathrm{I}\:\mathrm{cannot}\:\mathrm{solve}\:\mathrm{the}\:\mathrm{1}^{\mathrm{st}} \:\mathrm{one} \\ $$$$\mathrm{plus}\:\mathrm{I}\:\mathrm{somehow}\:\mathrm{solved}\:\mathrm{the}\:\mathrm{2}^{\mathrm{nd}} \:\mathrm{one}\:\mathrm{but}\:\mathrm{I}\:\mathrm{really} \\ $$$$\mathrm{don}'\mathrm{t}\:\mathrm{remember}\:\mathrm{how}…\:\mathrm{I}\:\mathrm{lost}\:\mathrm{some}\:\mathrm{papers}… \\ $$$$\mathrm{anyway}\:\frac{{d}}{{dt}}\left[\frac{\mathrm{4}}{\mathrm{21}}{t}\left(\mathrm{3}{t}^{\mathrm{2}} +\mathrm{1}\right)\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \right]=\mathrm{2}\frac{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{21}}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} } \\ $$$$\mathrm{and}\:−\frac{\mathrm{4}}{\mathrm{21}\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} }+\mathrm{2}\frac{{t}^{\mathrm{4}} +{t}^{\mathrm{2}} +\frac{\mathrm{2}}{\mathrm{21}}}{\left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{3}}{\mathrm{4}}} }=\mathrm{2}{t}^{\mathrm{2}} \left({t}^{\mathrm{2}} +\mathrm{1}\right)^{\frac{\mathrm{1}}{\mathrm{4}}} \\ $$
Commented by Abdo msup. last updated on 28/Aug/19
let take  a try   we do the changement (√(x+1))=t ⇒  x+1=t^2  ⇒x =t^2 −1 ⇒  ∫(√(x(√(x+1))))dx =∫  (√((t^2 −1)t))(2t)dt  =2 ∫  t(√(t^3 −t))dt =2∫ t(√(t(t^2 −1)))dt  =2 ∫ t^(3/2) (√(t^2 −1))dt   and changement t =ch(u) give  I =2 ∫   (chu)^(3/2) sh(u)sh(u)du  by psrts   f^′ =sh(u)(chu)^(3/2)   and g=shu  I =(2/5)(chu)^(5/2)  sh(u)−∫  (2/5)(chu)^(5/2)  ch(u)du  =(2/5)(chu)^(5/2) −(2/5) ∫  (chu)^(7/2)  du  the probleme is how to calculate intehral at  form ∫  (chu)^r du  with r ∈ Q^+ ?   ...be continued...
$${let}\:{take}\:\:{a}\:{try}\:\:\:{we}\:{do}\:{the}\:{changement}\:\sqrt{{x}+\mathrm{1}}={t}\:\Rightarrow \\ $$$${x}+\mathrm{1}={t}^{\mathrm{2}} \:\Rightarrow{x}\:={t}^{\mathrm{2}} −\mathrm{1}\:\Rightarrow \\ $$$$\int\sqrt{{x}\sqrt{{x}+\mathrm{1}}}{dx}\:=\int\:\:\sqrt{\left({t}^{\mathrm{2}} −\mathrm{1}\right){t}}\left(\mathrm{2}{t}\right){dt} \\ $$$$=\mathrm{2}\:\int\:\:{t}\sqrt{{t}^{\mathrm{3}} −{t}}{dt}\:=\mathrm{2}\int\:{t}\sqrt{{t}\left({t}^{\mathrm{2}} −\mathrm{1}\right)}{dt} \\ $$$$=\mathrm{2}\:\int\:{t}^{\frac{\mathrm{3}}{\mathrm{2}}} \sqrt{{t}^{\mathrm{2}} −\mathrm{1}}{dt}\:\:\:{and}\:{changement}\:{t}\:={ch}\left({u}\right)\:{give} \\ $$$${I}\:=\mathrm{2}\:\int\:\:\:\left({chu}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} {sh}\left({u}\right){sh}\left({u}\right){du} \\ $$$${by}\:{psrts}\:\:\:{f}^{'} ={sh}\left({u}\right)\left({chu}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \:\:{and}\:{g}={shu} \\ $$$${I}\:=\frac{\mathrm{2}}{\mathrm{5}}\left({chu}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:{sh}\left({u}\right)−\int\:\:\frac{\mathrm{2}}{\mathrm{5}}\left({chu}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} \:{ch}\left({u}\right){du} \\ $$$$=\frac{\mathrm{2}}{\mathrm{5}}\left({chu}\right)^{\frac{\mathrm{5}}{\mathrm{2}}} −\frac{\mathrm{2}}{\mathrm{5}}\:\int\:\:\left({chu}\right)^{\frac{\mathrm{7}}{\mathrm{2}}} \:{du} \\ $$$${the}\:{probleme}\:{is}\:{how}\:{to}\:{calculate}\:{intehral}\:{at} \\ $$$${form}\:\int\:\:\left({chu}\right)^{{r}} {du}\:\:{with}\:{r}\:\in\:{Q}^{+} ?\:\:\:…{be}\:{continued}… \\ $$
Commented by Abdo msup. last updated on 28/Aug/19
sir mjs  this integral is not  simple  to solve...
$${sir}\:{mjs}\:\:{this}\:{integral}\:{is}\:{not}\:\:{simple}\:\:{to}\:{solve}… \\ $$
Commented by MJS last updated on 28/Aug/19
thank you for trying
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{for}\:\mathrm{trying} \\ $$

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