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Question Number 128826 by benjo_mathlover last updated on 10/Jan/21

∫_(−1) ^( 5) (√((2x^2 −8)/x)) dx =?

$$\int_{−\mathrm{1}} ^{\:\mathrm{5}} \sqrt{\frac{\mathrm{2x}^{\mathrm{2}} −\mathrm{8}}{\mathrm{x}}}\:\mathrm{dx}\:=? \\ $$

Answered by bobhans last updated on 10/Jan/21

 ∫ (√(((x(√2))^2 −((√8))^2 )/x)) dx    let x(√2) = (√8) sec h   ∫ (√((8(sec^2 h−1))/(2sec h))) (2sec h tan h dh)  = 4 ∫ tan^2 h (√(sec h)) dh

$$\:\int\:\sqrt{\frac{\left({x}\sqrt{\mathrm{2}}\right)^{\mathrm{2}} −\left(\sqrt{\mathrm{8}}\right)^{\mathrm{2}} }{{x}}}\:{dx}\: \\ $$$$\:{let}\:{x}\sqrt{\mathrm{2}}\:=\:\sqrt{\mathrm{8}}\:\mathrm{sec}\:{h}\: \\ $$$$\int\:\sqrt{\frac{\mathrm{8}\left(\mathrm{sec}\:^{\mathrm{2}} {h}−\mathrm{1}\right)}{\mathrm{2sec}\:{h}}}\:\left(\mathrm{2sec}\:{h}\:\mathrm{tan}\:{h}\:{dh}\right) \\ $$$$=\:\mathrm{4}\:\int\:\mathrm{tan}\:^{\mathrm{2}} {h}\:\sqrt{\mathrm{sec}\:{h}}\:{dh}\: \\ $$$$ \\ $$

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