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Question Number 56392 by otchereabdullai@gmail.com last updated on 15/Mar/19

find the integral of ∫x(√(x−1))dx

$$\mathrm{find}\:\mathrm{the}\:\mathrm{integral}\:\mathrm{of}\:\:\int\mathrm{x}\sqrt{\mathrm{x}−\mathrm{1}}\mathrm{dx} \\ $$

Commented by maxmathsup by imad last updated on 16/Mar/19

changement (√(x−1))=t give x−1 =t^2 ⇒ ∫ x(√(x−1))dx =∫ (1+t^2 )t (2t)dt =2 ∫ (t^2 +t^4 )dt =(2/3)t^3 +(2/5)t^5 +c =(2/3)((√(x−1)))^3 +(2/5)((√(x−1))))^5 +c . =(2/3)(x−1)(√(x−1)) +(2/5)(x−1)^2 (√(x−1))+c .

$${changement}\:\sqrt{{x}−\mathrm{1}}={t}\:{give}\:{x}−\mathrm{1}\:={t}^{\mathrm{2}} \:\Rightarrow \\ $$$$\int\:{x}\sqrt{{x}−\mathrm{1}}{dx}\:=\int\:\left(\mathrm{1}+{t}^{\mathrm{2}} \right){t}\:\left(\mathrm{2}{t}\right){dt}\:=\mathrm{2}\:\int\:\left({t}^{\mathrm{2}} \:+{t}^{\mathrm{4}} \right){dt}\:=\frac{\mathrm{2}}{\mathrm{3}}{t}^{\mathrm{3}} \:+\frac{\mathrm{2}}{\mathrm{5}}{t}^{\mathrm{5}} \:+{c} \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left(\sqrt{{x}−\mathrm{1}}\right)^{\mathrm{3}} \:+\frac{\mathrm{2}}{\mathrm{5}}\left(\sqrt{\left.{x}−\mathrm{1}\right)}\right)^{\mathrm{5}} \:+{c}\:. \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\mathrm{1}\right)\sqrt{{x}−\mathrm{1}}\:+\frac{\mathrm{2}}{\mathrm{5}}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}−\mathrm{1}}+{c}\:. \\ $$

Commented by otchereabdullai@gmail.com last updated on 16/Mar/19

thank you sir!

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir}! \\ $$

Answered by kaivan.ahmadi last updated on 15/Mar/19

u=x−1⇒du=dx and x=u+1 ∫(u+1)(√u)du=∫u(√u)du+∫(√u)du= ∫u^(3/2) du+∫u^(1/2) du=((2u^(5/2) )/5)+((2u^(3/2) )/3)+C= ((2(x−1)^2 (√(x−1)))/5)+((2(x−1)(√(x−1)))/3)+C

$${u}={x}−\mathrm{1}\Rightarrow{du}={dx}\:{and}\:{x}={u}+\mathrm{1} \\ $$$$\int\left({u}+\mathrm{1}\right)\sqrt{{u}}{du}=\int{u}\sqrt{{u}}{du}+\int\sqrt{{u}}{du}= \\ $$$$\int{u}^{\frac{\mathrm{3}}{\mathrm{2}}} {du}+\int{u}^{\frac{\mathrm{1}}{\mathrm{2}}} {du}=\frac{\mathrm{2}{u}^{\frac{\mathrm{5}}{\mathrm{2}}} }{\mathrm{5}}+\frac{\mathrm{2}{u}^{\frac{\mathrm{3}}{\mathrm{2}}} }{\mathrm{3}}+{C}= \\ $$$$\frac{\mathrm{2}\left({x}−\mathrm{1}\right)^{\mathrm{2}} \sqrt{{x}−\mathrm{1}}}{\mathrm{5}}+\frac{\mathrm{2}\left({x}−\mathrm{1}\right)\sqrt{{x}−\mathrm{1}}}{\mathrm{3}}+{C} \\ $$

Commented by otchereabdullai@gmail.com last updated on 16/Mar/19

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

Answered by MJS last updated on 15/Mar/19

∫x(√(x−1))dx= [t=x−1 → dx=dt] =∫(t+1)(√t)dt=∫t^(3/2) dt+∫t^(1/2) dt=(2/5)t^(5/2) +(2/3)t^(3/2) = =(2/(15))t^(3/2) (3t+5)=(2/(15))(3x+2)(√((x−1)^3 ))+C

$$\int{x}\sqrt{{x}−\mathrm{1}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}={x}−\mathrm{1}\:\rightarrow\:{dx}={dt}\right] \\ $$$$=\int\left({t}+\mathrm{1}\right)\sqrt{{t}}{dt}=\int{t}^{\frac{\mathrm{3}}{\mathrm{2}}} {dt}+\int{t}^{\frac{\mathrm{1}}{\mathrm{2}}} {dt}=\frac{\mathrm{2}}{\mathrm{5}}{t}^{\frac{\mathrm{5}}{\mathrm{2}}} +\frac{\mathrm{2}}{\mathrm{3}}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} = \\ $$$$=\frac{\mathrm{2}}{\mathrm{15}}{t}^{\frac{\mathrm{3}}{\mathrm{2}}} \left(\mathrm{3}{t}+\mathrm{5}\right)=\frac{\mathrm{2}}{\mathrm{15}}\left(\mathrm{3}{x}+\mathrm{2}\right)\sqrt{\left({x}−\mathrm{1}\right)^{\mathrm{3}} }+{C} \\ $$

Commented by otchereabdullai@gmail.com last updated on 16/Mar/19

thank you sir

$$\mathrm{thank}\:\mathrm{you}\:\mathrm{sir} \\ $$

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