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I-x-x-2-1-dx-




Question Number 31839 by Joel578 last updated on 15/Mar/18
I = ∫ (√(x + (√(x^2  − 1)))) dx
$${I}\:=\:\int\:\sqrt{{x}\:+\:\sqrt{{x}^{\mathrm{2}} \:−\:\mathrm{1}}}\:{dx} \\ $$
Commented by Joel578 last updated on 15/Mar/18
   = ∫ (√(x + (√(4 . ((x^2  − 1)/4))))) dx     = ∫ (√(((x + 1)/2) + ((x − 1)/2) + 2(√(((x +1)(x − 1))/(2 . 2))))) dx     = ∫ (√((x + 1)/2)) + (√((x − 1)/2)) dx     = (1/( (√2))) ∫ (√(x + 1)) + (√(x − 1)) dx     = (1/( (√2))) ((2/3)(x + 1)^(3/2)  + (2/3)(x − 1)^(3/2) ) + C     = ((√2)/3)[(x + 1)^(3/2)  + (x − 1)^(3/2) ) + C
$$\:\:\:=\:\int\:\sqrt{{x}\:+\:\sqrt{\mathrm{4}\:.\:\frac{{x}^{\mathrm{2}} \:−\:\mathrm{1}}{\mathrm{4}}}}\:{dx} \\ $$$$\:\:\:=\:\int\:\sqrt{\frac{{x}\:+\:\mathrm{1}}{\mathrm{2}}\:+\:\frac{{x}\:−\:\mathrm{1}}{\mathrm{2}}\:+\:\mathrm{2}\sqrt{\frac{\left({x}\:+\mathrm{1}\right)\left({x}\:−\:\mathrm{1}\right)}{\mathrm{2}\:.\:\mathrm{2}}}}\:{dx} \\ $$$$\:\:\:=\:\int\:\sqrt{\frac{{x}\:+\:\mathrm{1}}{\mathrm{2}}}\:+\:\sqrt{\frac{{x}\:−\:\mathrm{1}}{\mathrm{2}}}\:{dx} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\int\:\sqrt{{x}\:+\:\mathrm{1}}\:+\:\sqrt{{x}\:−\:\mathrm{1}}\:{dx} \\ $$$$\:\:\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\left(\frac{\mathrm{2}}{\mathrm{3}}\left({x}\:+\:\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} \:+\:\frac{\mathrm{2}}{\mathrm{3}}\left({x}\:−\:\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} \right)\:+\:{C} \\ $$$$\:\:\:=\:\frac{\sqrt{\mathrm{2}}}{\mathrm{3}}\left[\left({x}\:+\:\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} \:+\:\left({x}\:−\:\mathrm{1}\right)^{\mathrm{3}/\mathrm{2}} \right)\:+\:{C} \\ $$

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