Question Number 46594 by maxmathsup by imad last updated on 29/Oct/18
$${find}\:\int\:\left(\sqrt{{x}+\sqrt{{x}}}−\sqrt{{x}−\sqrt{{x}}}\right){dx} \\ $$
Answered by MJS last updated on 29/Oct/18
![∫(√(x+(√x)))dx= [t=(√x) → dx=2(√x)dt] =∫2t(√(t^2 +t))dt=∫(2t+1)(√(t^2 +t))dt−∫(√(t^2 +t))dt ∫(2t+1)(√(t^2 +t))dt=(2/3)(t^2 +t)^(3/2) =(2/3)(x+(√x))^(3/2) −∫(√(t^2 +t))dt=−(1/2)∫(√((2t+1)^2 −1))dt= [u=2t+1 → dt=(1/2)du] =−(1/4)∫(√(u^2 −1))du= [v=arccosh u → du=sinh v dv] =−(1/4)∫sinh^2 v dv=(1/8)∫dv−(1/8)∫cosh 2v dv= =(1/8)v−(1/(16))sinh 2v =(1/8)arccosh u −(1/(16))sinh (2arccosh u)= =(1/8)arccosh (2t+1) −(1/(16))sinh (2arccosh (2t+1))= =(1/8)arccosh (2(√x)+1) −(1/(16))sinh (2arccosh (2(√x)+1)) ∫(√(x+(√x)))dx=(2/3)(x+(√x))^(3/2) +(1/8)arccosh (2(√x)+1) −(1/(16))sinh (2arccosh (2(√x)+1)) +C similar −∫(√(x−(√x)))dx=−(2/3)(x−(√x))^(3/2) +(1/8)arccosh (2(√x)−1) −(1/(16))sinh (2arccosh (2(√x)−1)) +C](Q46601.png)
$$\int\sqrt{{x}+\sqrt{{x}}}{dx}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}}{dt}\right] \\ $$$$=\int\mathrm{2}{t}\sqrt{{t}^{\mathrm{2}} +{t}}{dt}=\int\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} +{t}}{dt}−\int\sqrt{{t}^{\mathrm{2}} +{t}}{dt} \\ $$$$ \\ $$$$\:\:\:\:\:\int\left(\mathrm{2}{t}+\mathrm{1}\right)\sqrt{{t}^{\mathrm{2}} +{t}}{dt}=\frac{\mathrm{2}}{\mathrm{3}}\left({t}^{\mathrm{2}} +{t}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} =\frac{\mathrm{2}}{\mathrm{3}}\left({x}+\sqrt{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} \\ $$$$ \\ $$$$\:\:\:\:\:−\int\sqrt{{t}^{\mathrm{2}} +{t}}{dt}=−\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{\left(\mathrm{2}{t}+\mathrm{1}\right)^{\mathrm{2}} −\mathrm{1}}{dt}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{u}=\mathrm{2}{t}+\mathrm{1}\:\rightarrow\:{dt}=\frac{\mathrm{1}}{\mathrm{2}}{du}\right] \\ $$$$\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{4}}\int\sqrt{{u}^{\mathrm{2}} −\mathrm{1}}{du}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{v}=\mathrm{arccosh}\:{u}\:\rightarrow\:{du}=\mathrm{sinh}\:{v}\:{dv}\right] \\ $$$$\:\:\:\:\:=−\frac{\mathrm{1}}{\mathrm{4}}\int\mathrm{sinh}^{\mathrm{2}} \:{v}\:{dv}=\frac{\mathrm{1}}{\mathrm{8}}\int{dv}−\frac{\mathrm{1}}{\mathrm{8}}\int\mathrm{cosh}\:\mathrm{2}{v}\:{dv}= \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8}}{v}−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sinh}\:\mathrm{2}{v}\:=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{arccosh}\:{u}\:−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sinh}\:\left(\mathrm{2arccosh}\:{u}\right)= \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{arccosh}\:\left(\mathrm{2}{t}+\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sinh}\:\left(\mathrm{2arccosh}\:\left(\mathrm{2t}+\mathrm{1}\right)\right)= \\ $$$$\:\:\:\:\:=\frac{\mathrm{1}}{\mathrm{8}}\mathrm{arccosh}\:\left(\mathrm{2}\sqrt{{x}}+\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sinh}\:\left(\mathrm{2arccosh}\:\left(\mathrm{2}\sqrt{{x}}+\mathrm{1}\right)\right) \\ $$$$ \\ $$$$\int\sqrt{{x}+\sqrt{{x}}}{dx}=\frac{\mathrm{2}}{\mathrm{3}}\left({x}+\sqrt{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{8}}\mathrm{arccosh}\:\left(\mathrm{2}\sqrt{{x}}+\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sinh}\:\left(\mathrm{2arccosh}\:\left(\mathrm{2}\sqrt{{x}}+\mathrm{1}\right)\right)\:+{C} \\ $$$$\mathrm{similar} \\ $$$$−\int\sqrt{{x}−\sqrt{{x}}}{dx}=−\frac{\mathrm{2}}{\mathrm{3}}\left({x}−\sqrt{{x}}\right)^{\frac{\mathrm{3}}{\mathrm{2}}} +\frac{\mathrm{1}}{\mathrm{8}}\mathrm{arccosh}\:\left(\mathrm{2}\sqrt{{x}}−\mathrm{1}\right)\:−\frac{\mathrm{1}}{\mathrm{16}}\mathrm{sinh}\:\left(\mathrm{2arccosh}\:\left(\mathrm{2}\sqrt{{x}}−\mathrm{1}\right)\right)\:+{C} \\ $$
Commented by maxmathsup by imad last updated on 29/Oct/18
$${good}\:{work}\:{is}\:{done}\:{thank}\:{you}\:{sirMJS}. \\ $$