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Question Number 147784 by Khalmohmmad last updated on 23/Jul/21

Answered by Olaf_Thorendsen last updated on 23/Jul/21

F(x) = ∫(dx/( (√(7x^2 −8)))) F(x) = (1/( (√7)))∫(dx/( (√(x^2 −(8/7))))) F(x) = (1/( (√7)))argch((x/( (√(8/( 7))))))+C F(x) = (1/( (√7)))argch((1/2)(√(7/2))x)+C

$${F}\left({x}\right)\:=\:\int\frac{{dx}}{\:\sqrt{\mathrm{7}{x}^{\mathrm{2}} −\mathrm{8}}} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\int\frac{{dx}}{\:\sqrt{{x}^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{7}}}} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\mathrm{argch}\left(\frac{{x}}{\:\sqrt{\frac{\mathrm{8}}{\:\mathrm{7}}}}\right)+\mathrm{C} \\ $$$${F}\left({x}\right)\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\mathrm{argch}\left(\frac{\mathrm{1}}{\mathrm{2}}\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}{x}\right)+\mathrm{C} \\ $$

Commented by mathmax by abdo last updated on 23/Jul/21

Υ=∫ (dx/( (√(7x^2 −8)))) ⇒Υ=(1/( (√7)))∫ (dx/( (√(x^2 −(8/7))))) =_(x=((2(√2))/( (√7)))ch(t)) (1/( (√7)))∫ (1/(((2(√2))/( (√7)))sh(t)))×((2(√2))/( (√7))) sht dt =(1/( (√7)))∫ dt =(t/( (√7))) +C but t=argch(((√7)/(2(√2)))x) =ln(((√7)/(2(√2)))x+(√(((7x^2 )/8)−1))) ⇒Υ=(1/( (√7)))ln((((√7)x)/(2(√2)))+(√(((7x^2 )/8)−1))) +C

$$\Upsilon=\int\:\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{7x}^{\mathrm{2}} −\mathrm{8}}}\:\Rightarrow\Upsilon=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\int\:\:\frac{\mathrm{dx}}{\:\sqrt{\mathrm{x}^{\mathrm{2}} −\frac{\mathrm{8}}{\mathrm{7}}}} \\ $$$$=_{\mathrm{x}=\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{7}}}\mathrm{ch}\left(\mathrm{t}\right)} \:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\int\:\:\frac{\mathrm{1}}{\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{7}}}\mathrm{sh}\left(\mathrm{t}\right)}×\frac{\mathrm{2}\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{7}}}\:\mathrm{sht}\:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\int\:\mathrm{dt}\:=\frac{\mathrm{t}}{\:\sqrt{\mathrm{7}}}\:+\mathrm{C}\:\:\:\:\mathrm{but}\:\mathrm{t}=\mathrm{argch}\left(\frac{\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{x}\right) \\ $$$$=\mathrm{ln}\left(\frac{\sqrt{\mathrm{7}}}{\mathrm{2}\sqrt{\mathrm{2}}}\mathrm{x}+\sqrt{\frac{\mathrm{7x}^{\mathrm{2}} }{\mathrm{8}}−\mathrm{1}}\right)\:\Rightarrow\Upsilon=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\mathrm{ln}\left(\frac{\sqrt{\mathrm{7}}\mathrm{x}}{\mathrm{2}\sqrt{\mathrm{2}}}+\sqrt{\frac{\mathrm{7x}^{\mathrm{2}} }{\mathrm{8}}−\mathrm{1}}\right)\:+\mathrm{C} \\ $$

Answered by iloveisrael last updated on 23/Jul/21

(√7) x =(√8) sec t ⇒cos t = ((√8)/( (√7) x)) ∫ ((((√8)/( (√7))) sec t tan t dt)/( (√(8sec^2 t−8)))) = (1/( (√7))) ∫ ((sec t tan t)/(tan t)) dt =(1/( (√7))) ln ∣sec t +tan t ∣ + c =(1/( (√7))) ln ∣(((√7) x)/( (√8))) +((√(7x^2 −8))/( (√8))) ∣ + c =(1/( (√7))) ln ∣ ((x(√7) +(√(7x^2 −8)))/( (√8))) ∣ + c

$$\sqrt{\mathrm{7}}\:\mathrm{x}\:=\sqrt{\mathrm{8}}\:\mathrm{sec}\:\mathrm{t}\:\Rightarrow\mathrm{cos}\:\mathrm{t}\:=\:\frac{\sqrt{\mathrm{8}}}{\:\sqrt{\mathrm{7}}\:\mathrm{x}} \\ $$$$\int\:\frac{\frac{\sqrt{\mathrm{8}}}{\:\sqrt{\mathrm{7}}}\:\mathrm{sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}\:\mathrm{dt}}{\:\sqrt{\mathrm{8sec}\:^{\mathrm{2}} \mathrm{t}−\mathrm{8}}}\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\:\int\:\frac{\mathrm{sec}\:\mathrm{t}\:\mathrm{tan}\:\mathrm{t}}{\mathrm{tan}\:\mathrm{t}}\:\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\:\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{t}\:+\mathrm{tan}\:\mathrm{t}\:\mid\:+\:\mathrm{c} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\:\mathrm{ln}\:\mid\frac{\sqrt{\mathrm{7}}\:\mathrm{x}}{\:\sqrt{\mathrm{8}}}\:+\frac{\sqrt{\mathrm{7x}^{\mathrm{2}} −\mathrm{8}}}{\:\sqrt{\mathrm{8}}}\:\mid\:+\:\mathrm{c} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\:\mathrm{ln}\:\mid\:\frac{\mathrm{x}\sqrt{\mathrm{7}}\:+\sqrt{\mathrm{7x}^{\mathrm{2}} −\mathrm{8}}}{\:\sqrt{\mathrm{8}}}\:\mid\:+\:\mathrm{c}\: \\ $$

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