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Question Number 73482 by abdomathmax last updated on 13/Nov/19

find ∫ (dx/((√(x^2 +1))+(√(x^2 +3))))

$${find}\:\int\:\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}} \\ $$

Commented by abdomathmax last updated on 17/Nov/19

let I =∫ (dx/((√(x^2 +1))+(√(x^2 +3)))) ⇒ I = ∫(((√(x^2 +3))−(√(x^2 +1)))/2)dx =(1/2)∫ (√(x^2 +3))dx−(1/2)∫(√(x^2 +1))dx we have ∫(√(x^2 +3))dx =_(x=(√3)sh(t)) (√3) ∫ (√3)ch(t)ch(t) =3∫ ((1+ch(2t))/2)dt =(3/2)t +(3/4)sh(2t) +c_1 =(3/4)t +(3/4)2sh(t)ch(t) +c_1 =(3/4) argsh((x/(√3))) +(3/2)(x/(√3))(√(1+(x^2 /3))) +c =(3/4)ln((x/(√3))+(√(1+(x^2 /3)))) +(x/2)(√(3+x^2 )) +c_1 ∫ (√(x^2 +1))dx =_(x=sh(t)) ∫ch(t)ch(t)dt =∫ ((1+ch(2t))/2)dt =(t/2) +(1/4)sh(2t) +c_2 =(t/2) +(1/2)sh(g)ch(t) +c_2 =(1/2)ln(x+(√(1+x^2 )))+(x/2)(√(1+x^2 )) +c_2 ⇒ I =(3/8)ln(x+(√(3+x^2 ))) +(x/4)(√(3+x^2 ))−(1/4)ln(x+(√(1+x^2 ))) −(x/4)(√(1+x^2 )) +C

$${let}\:{I}\:=\int\:\:\:\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} \:+\mathrm{3}}}\:\Rightarrow \\ $$$${I}\:=\:\int\frac{\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}{\mathrm{2}}{dx}\:=\frac{\mathrm{1}}{\mathrm{2}}\int\:\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${we}\:{have}\:\int\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}{dx}\:=_{{x}=\sqrt{\mathrm{3}}{sh}\left({t}\right)} \:\sqrt{\mathrm{3}}\:\int\:\sqrt{\mathrm{3}}{ch}\left({t}\right){ch}\left({t}\right) \\ $$$$=\mathrm{3}\int\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}\:=\frac{\mathrm{3}}{\mathrm{2}}{t}\:+\frac{\mathrm{3}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:+{c}_{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{t}\:\:+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{2}{sh}\left({t}\right){ch}\left({t}\right)\:+{c}_{\mathrm{1}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:{argsh}\left(\frac{{x}}{\sqrt{\mathrm{3}}}\right)\:+\frac{\mathrm{3}}{\mathrm{2}}\frac{{x}}{\sqrt{\mathrm{3}}}\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}\:+{c} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}{ln}\left(\frac{{x}}{\sqrt{\mathrm{3}}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{3}}}\right)\:+\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }\:+{c}_{\mathrm{1}} \\ $$$$\int\:\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}{dx}\:=_{{x}={sh}\left({t}\right)} \:\:\:\int{ch}\left({t}\right){ch}\left({t}\right){dt} \\ $$$$=\int\:\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}{dt}\:=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{4}}{sh}\left(\mathrm{2}{t}\right)\:+{c}_{\mathrm{2}} \\ $$$$=\frac{{t}}{\mathrm{2}}\:+\frac{\mathrm{1}}{\mathrm{2}}{sh}\left({g}\right){ch}\left({t}\right)\:+{c}_{\mathrm{2}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right)+\frac{{x}}{\mathrm{2}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+{c}_{\mathrm{2}} \:\Rightarrow \\ $$$${I}\:=\frac{\mathrm{3}}{\mathrm{8}}{ln}\left({x}+\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }\right)\:+\frac{{x}}{\mathrm{4}}\sqrt{\mathrm{3}+{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{\mathrm{4}}{ln}\left({x}+\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\right) \\ $$$$−\frac{{x}}{\mathrm{4}}\sqrt{\mathrm{1}+{x}^{\mathrm{2}} }\:+{C} \\ $$$$ \\ $$

Answered by ajfour last updated on 13/Nov/19

I=∫(dx/((√(x^2 +1))+(√(x^2 +3)))) =(1/2)∫((√(x^2 +3))−(√(x^2 +1)) )dx =(x/4)((√(x^2 +3))−(√(x^2 +1)) ) +(3/4)ln ∣x+(√(x^2 +3))∣−(1/4)ln ∣x+(√(x^2 +1)) ∣+c __________________________.

$${I}=\int\frac{{dx}}{\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\right){dx} \\ $$$$\:\:\:=\frac{{x}}{\mathrm{4}}\left(\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}−\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\right) \\ $$$$+\frac{\mathrm{3}}{\mathrm{4}}\mathrm{ln}\:\mid{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{3}}\mid−\frac{\mathrm{1}}{\mathrm{4}}\mathrm{ln}\:\mid{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}\:\mid+{c} \\ $$$$\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_\_. \\ $$

Commented by abdomathmax last updated on 17/Nov/19

thanks sir.

$${thanks}\:{sir}. \\ $$

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