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Question Number 86092 by ar247 last updated on 27/Mar/20

∫(1/(√(5−4x−2x^2 )))dx

$$\int\frac{\mathrm{1}}{\sqrt{\mathrm{5}−\mathrm{4}{x}−\mathrm{2}{x}^{\mathrm{2}} }}{dx} \\ $$

Commented by abdomathmax last updated on 27/Mar/20

I =∫ (dx/(√(−2x^2 −4x+5))) we have I=∫ (dx/(√(5−2(x^2 +2x+1−1))))=∫ (dx/(√(5−2(x+1)^2 +2))) =∫ (dx/(√(7−2(x+1)^2 ))) =∫ (dx/((√7)×(√(1−(2/7)(x+1)^2 )))) vhangement (√(2/7))(x+1)=u give I =(1/(√7)) ∫ (1/(√(1−u^2 )))×((√7)/(√2))du =(1/(√2)) arcsin(u) +C =(1/(√2)) arcsin((√(2/7))(x+1)) +C

$${I}\:=\int\:\:\frac{{dx}}{\sqrt{−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}}}\:\:{we}\:{have} \\ $$$${I}=\int\:\:\:\frac{{dx}}{\sqrt{\mathrm{5}−\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}−\mathrm{1}\right)}}=\int\:\:\frac{{dx}}{\sqrt{\mathrm{5}−\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}}} \\ $$$$=\int\:\:\:\frac{{dx}}{\sqrt{\mathrm{7}−\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }}\:=\int\:\:\frac{{dx}}{\sqrt{\mathrm{7}}×\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{7}}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$${vhangement}\:\:\sqrt{\frac{\mathrm{2}}{\mathrm{7}}}\left({x}+\mathrm{1}\right)={u}\:{give} \\ $$$${I}\:=\frac{\mathrm{1}}{\sqrt{\mathrm{7}}}\:\int\:\:\:\frac{\mathrm{1}}{\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}×\frac{\sqrt{\mathrm{7}}}{\sqrt{\mathrm{2}}}{du} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\:{arcsin}\left({u}\right)\:+{C} \\ $$$$=\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:{arcsin}\left(\sqrt{\frac{\mathrm{2}}{\mathrm{7}}}\left({x}+\mathrm{1}\right)\right)\:+{C} \\ $$

Answered by jagoll last updated on 27/Mar/20

1+4−4x−2x^2 = 1+2(2−2x−x^2 ) 1+2(2−(x^2 +2x+1)) = 1+2(3−(x+1)^2 ) =7−2(x+1)^2 let x+1 = (√(7/2)) sin t ∫ (((√(7/2)) cos t dt)/(√(7−7sin^2 t))) = ((√2)/2)∫ dt = (1/2) t +c ((√2)/2) arc sin (((√2)/(√7)) (x+1)) +c

$$\mathrm{1}+\mathrm{4}−\mathrm{4x}−\mathrm{2x}^{\mathrm{2}} \:=\:\mathrm{1}+\mathrm{2}\left(\mathrm{2}−\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\mathrm{1}+\mathrm{2}\left(\mathrm{2}−\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}\right)\right)\:= \\ $$$$\mathrm{1}+\mathrm{2}\left(\mathrm{3}−\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \right)\:=\mathrm{7}−\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{x}+\mathrm{1}\:=\:\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\mathrm{sin}\:\mathrm{t} \\ $$$$\int\:\frac{\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\mathrm{cos}\:\mathrm{t}\:\mathrm{dt}}{\sqrt{\mathrm{7}−\mathrm{7sin}\:^{\mathrm{2}} \mathrm{t}}}\:=\: \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\:\mathrm{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{t}\:+\mathrm{c}\: \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{arc}\:\mathrm{sin}\:\left(\frac{\sqrt{\mathrm{2}}}{\sqrt{\mathrm{7}}}\:\left(\mathrm{x}+\mathrm{1}\right)\right)\:+\mathrm{c} \\ $$$$ \\ $$

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