Question Number 115594 by MJS_new last updated on 26/Sep/20
$$\mathrm{old}\:\mathrm{question},\:\mathrm{I}\:\mathrm{couldn}'\mathrm{t}\:\mathrm{find}\:\mathrm{it}: \\ $$$$\int\sqrt{{x}−\sqrt{{x}}}{dx}=? \\ $$
Answered by MJS_new last updated on 26/Sep/20
![∫(√(x−(√x)))dx= [t=(√x) → dx=2(√x)dt] =2∫t^(3/2) (√(t−1))dt= [u=(√(t−1)) → dt=2(√(t−1))du] =4∫u^2 (u^2 +1)^(3/2) du= [v=u+(√(u^2 +1)) → du=((√(u^2 +1))/(u+(√(u^2 +1))))dv] =(1/(16))∫((v^(12) +2v^(10) −v^8 −4v^6 −v^4 +2v^2 +1)/v^7 )dv= =(v^6 /(96))+(v^4 /(32))−(v^2 /(32))+(1/(32v^2 ))−(1/(32v^4 ))−(1/(96v^6 ))−(1/4)ln v = [v=u+(√(u^2 +1))∧u=(√(t−1))∧t=(√x)] =(1/(12))(8x−2(√x)−3)(√(x−(√x)))−(1/4)ln ((√((√x)−1))+(x)^(1/4) ) +C](Q115596.png)
$$\int\sqrt{{x}−\sqrt{{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\ $$$$=\mathrm{2}\int{t}^{\mathrm{3}/\mathrm{2}} \sqrt{{t}−\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\left[{u}=\sqrt{{t}−\mathrm{1}}\:\rightarrow\:{dt}=\mathrm{2}\sqrt{{t}−\mathrm{1}}{du}\right] \\ $$$$=\mathrm{4}\int{u}^{\mathrm{2}} \left({u}^{\mathrm{2}} +\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} {du}= \\ $$$$\:\:\:\:\:\left[{v}={u}+\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}\:\rightarrow\:{du}=\frac{\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}{{u}+\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}}{dv}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{16}}\int\frac{{v}^{\mathrm{12}} +\mathrm{2}{v}^{\mathrm{10}} −{v}^{\mathrm{8}} −\mathrm{4}{v}^{\mathrm{6}} −{v}^{\mathrm{4}} +\mathrm{2}{v}^{\mathrm{2}} +\mathrm{1}}{{v}^{\mathrm{7}} }{dv}= \\ $$$$=\frac{{v}^{\mathrm{6}} }{\mathrm{96}}+\frac{{v}^{\mathrm{4}} }{\mathrm{32}}−\frac{{v}^{\mathrm{2}} }{\mathrm{32}}+\frac{\mathrm{1}}{\mathrm{32}{v}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{32}{v}^{\mathrm{4}} }−\frac{\mathrm{1}}{\mathrm{96}{v}^{\mathrm{6}} }−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:{v}\:= \\ $$$$\:\:\:\:\:\left[{v}={u}+\sqrt{{u}^{\mathrm{2}} +\mathrm{1}}\wedge{u}=\sqrt{{t}−\mathrm{1}}\wedge{t}=\sqrt{{x}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{12}}\left(\mathrm{8}{x}−\mathrm{2}\sqrt{{x}}−\mathrm{3}\right)\sqrt{{x}−\sqrt{{x}}}−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\left(\sqrt{\sqrt{{x}}−\mathrm{1}}+\sqrt[{\mathrm{4}}]{{x}}\right)\:+{C} \\ $$
Commented by Eric002 last updated on 27/Sep/20
$${nice}\:{work}\:{i}\:{think}\:{its}\:{q}\mathrm{115498} \\ $$