Question Number 86092 by ar247 last updated on 27/Mar/20
$$\int\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}−\mathrm{4}{x}−\mathrm{2}{x}^{\mathrm{2}} }}{dx} \\ $$
Commented by abdomathmax last updated on 27/Mar/20
$${I}\:=\int\:\:\frac{{dx}}{\:\sqrt{−\mathrm{2}{x}^{\mathrm{2}} −\mathrm{4}{x}+\mathrm{5}}}\:\:{we}\:{have} \\ $$$${I}=\int\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{5}−\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}{x}+\mathrm{1}−\mathrm{1}\right)}}=\int\:\:\frac{{dx}}{\:\sqrt{\mathrm{5}−\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}}} \\ $$$$=\int\:\:\:\frac{{dx}}{\:\sqrt{\mathrm{7}−\mathrm{2}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }}\:=\int\:\:\frac{{dx}}{\:\sqrt{\mathrm{7}}×\sqrt{\mathrm{1}−\frac{\mathrm{2}}{\mathrm{7}}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }} \\ $$$${vhangement}\:\:\sqrt{\frac{\mathrm{2}}{\mathrm{7}}}\left({x}+\mathrm{1}\right)={u}\:{give} \\ $$$${I}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{7}}}\:\int\:\:\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{1}−{u}^{\mathrm{2}} }}×\frac{\sqrt{\mathrm{7}}}{\:\sqrt{\mathrm{2}}}{du} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:\:{arcsin}\left({u}\right)\:+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}}\:{arcsin}\left(\sqrt{\frac{\mathrm{2}}{\mathrm{7}}}\left({x}+\mathrm{1}\right)\right)\:+{C} \\ $$
Answered by jagoll last updated on 27/Mar/20
$$\mathrm{1}+\mathrm{4}−\mathrm{4x}−\mathrm{2x}^{\mathrm{2}} \:=\:\mathrm{1}+\mathrm{2}\left(\mathrm{2}−\mathrm{2x}−\mathrm{x}^{\mathrm{2}} \right) \\ $$$$\mathrm{1}+\mathrm{2}\left(\mathrm{2}−\left(\mathrm{x}^{\mathrm{2}} +\mathrm{2x}+\mathrm{1}\right)\right)\:= \\ $$$$\mathrm{1}+\mathrm{2}\left(\mathrm{3}−\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \right)\:=\mathrm{7}−\mathrm{2}\left(\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} \\ $$$$\mathrm{let}\:\mathrm{x}+\mathrm{1}\:=\:\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\mathrm{sin}\:\mathrm{t} \\ $$$$\int\:\frac{\sqrt{\frac{\mathrm{7}}{\mathrm{2}}}\:\mathrm{cos}\:\mathrm{t}\:\mathrm{dt}}{\:\sqrt{\mathrm{7}−\mathrm{7sin}\:^{\mathrm{2}} \mathrm{t}}}\:=\: \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\int\:\mathrm{dt}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{t}\:+\mathrm{c}\: \\ $$$$\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\:\mathrm{arc}\:\mathrm{sin}\:\left(\frac{\sqrt{\mathrm{2}}}{\:\sqrt{\mathrm{7}}}\:\left(\mathrm{x}+\mathrm{1}\right)\right)\:+\mathrm{c} \\ $$$$ \\ $$