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Question Number 96175 by Fikret last updated on 30/May/20

∫(dx/(√(4x^2 +4x+3)))=?

$$\int\frac{{dx}}{\sqrt{\mathrm{4}{x}^{\mathrm{2}} +\mathrm{4}{x}+\mathrm{3}}}=? \\ $$

Commented by bobhans last updated on 30/May/20

∫ (dx/(2(√(x^2 +x+(3/4))))) = ∫ (dx/(2(√((x+(1/2))^2 +(1/2)))))  set x+(1/2) = (√(1/2)) tan u ⇒tan u = ((2x+1)/(√2))  =(1/2)∫ (((1/(√2)) sec^2 u du)/((1/(√2)) sec u)) = (1/2)ln ∣sec u+tan u∣ +c

$$\int\:\frac{{dx}}{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\frac{\mathrm{3}}{\mathrm{4}}}}\:=\:\int\:\frac{{dx}}{\mathrm{2}\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}}} \\ $$$${set}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}\:=\:\sqrt{\frac{\mathrm{1}}{\mathrm{2}}}\:\mathrm{tan}\:{u}\:\Rightarrow\mathrm{tan}\:\mathrm{u}\:=\:\frac{\mathrm{2x}+\mathrm{1}}{\sqrt{\mathrm{2}}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{sec}\:^{\mathrm{2}} \mathrm{u}\:\mathrm{du}}{\frac{\mathrm{1}}{\sqrt{\mathrm{2}}}\:\mathrm{sec}\:\mathrm{u}}\:=\:\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\mid\mathrm{sec}\:\mathrm{u}+\mathrm{tan}\:\mathrm{u}\mid\:+\mathrm{c} \\ $$$$ \\ $$

Answered by mathmax by abdo last updated on 30/May/20

I =∫  (dx/(√(4x^2  +4x+3))) ⇒I =∫  (dx/(√((2x)^2 +2(2x)+1 +2)))  =∫  (dx/(√((2x+1)^2 +2))) =_(2x+1 =(√2)u)      ∫  (du/((√2)(√2)(√(1+u^2 )))) =(1/2) argsh(u) +c  =(1/2)ln(u+(√(1+u^2 ))) +c =(1/2)ln(((2x+1)/(√2)) +(√(1+(((2x+1)/(√2)))^2 ))) +c

$$\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\sqrt{\mathrm{4x}^{\mathrm{2}} \:+\mathrm{4x}+\mathrm{3}}}\:\Rightarrow\mathrm{I}\:=\int\:\:\frac{\mathrm{dx}}{\sqrt{\left(\mathrm{2x}\right)^{\mathrm{2}} +\mathrm{2}\left(\mathrm{2x}\right)+\mathrm{1}\:+\mathrm{2}}} \\ $$$$=\int\:\:\frac{\mathrm{dx}}{\sqrt{\left(\mathrm{2x}+\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}}}\:=_{\mathrm{2x}+\mathrm{1}\:=\sqrt{\mathrm{2}}\mathrm{u}} \:\:\:\:\:\int\:\:\frac{\mathrm{du}}{\sqrt{\mathrm{2}}\sqrt{\mathrm{2}}\sqrt{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }}\:=\frac{\mathrm{1}}{\mathrm{2}}\:\mathrm{argsh}\left(\mathrm{u}\right)\:+\mathrm{c} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\mathrm{u}+\sqrt{\mathrm{1}+\mathrm{u}^{\mathrm{2}} }\right)\:+\mathrm{c}\:=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left(\frac{\mathrm{2x}+\mathrm{1}}{\sqrt{\mathrm{2}}}\:+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2x}+\mathrm{1}}{\sqrt{\mathrm{2}}}\right)^{\mathrm{2}} }\right)\:+\mathrm{c} \\ $$

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