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Question Number 73144 by mathmax by abdo last updated on 06/Nov/19
calculte ∫ ((x+(√(2+x^2 )))/(x+1−(√(2+x^2 ))))dx
$${calculte}\:\int\:\:\frac{{x}+\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{{x}+\mathrm{1}−\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{dx} \\ $$
Commented by mathmax by abdo last updated on 07/Nov/19
I=∫ ((x+(√(2+x^2 )))/(x+1−(√(2+x^2 ))))dx changement x=(√2)sh(t) give I =∫ (((√2)sh(t)+(√2)ch(t))/( (√2)sh(t)+1−(√2)cht))(√2)cht dt = ∫ (((2sh(t)ch(t)+ch^2 t))/( (√2)sh(t)−(√2)ch(t)+1))dt =∫ ((sh(2t)+((1+ch(2t))/2))/( (√2)sh(t)−(√2)ch(t)+1))dt =∫ ((2sh(2t)+ch(2t)+1)/(2((√2)sh(t)−(√2)ch(t) +1)))dt =(1/2)∫ ((2((e^(2t) −e^(−2t) )/2)+((e^(2t) +e^(−2t) )/2) +1)/( (√2)((e^t −e^(−t) )/2)−(√2)((e^t +e^(−t) )/2)+1))dt =(1/2) ∫ ((2e^(2t) −2e^(−2t) +e^(2t) +e^(−2t) +2)/( (√2)e^t −(√2)e^(−t) −(√2)e^t −(√2)e^(−t) +2))dt ⇒ 2I =∫ ((3e^(2t) −e^(−2t) +2)/(−2(√2)e^(−t) +2))dt =_(e^t =u) ∫ ((3u^2 −u^(−2) +2)/(2−2(√2)u^(−1) ))(du/u) =∫ ((3u^2 −u^(−2) )/(2u−2(√2)))du =∫ ((3u^4 −1)/(2u^3 −2(√2)u))du ⇒ 4I = ∫ ((3u^4 −1)/(u^3 −(√2)u))du =∫ ((3u(u^3 −(√2)u)+3(√2)u^2 −1)/(u^3 −(√2)u))du =3u +∫ ((3(√2)u^2 −1)/(u(u^2 −(√2))))du let decompose F(u)=((3(√2)u^2 −1)/(u(u^2 −(√2)))) ⇒F(u) =((3(√2)u^2 −1)/(u(u−α)(u+α))) with α=^4 (√2) F(u) =(a/u) +(b/(u−α)) +(c/(u+α)) a =((−1)/(−α^2 )) =(1/α^2 ) =(1/( (√2))) b =((3(√2)α^2 −1)/(α(2α))) =((3(√2)α^2 −1)/(2α^2 )) and c =((3(√2)α^2 −1)/((−α)(−2α))) =((3(√2)α^2 −1)/(2α^2 )) and ∫ F(u)du =aln∣u∣+bln∣u−α∣+cln∣u+α∣ c =at +bln∣e^t −α∣ +cln∣e^t +α∣ +c but t=argsh((x/( (√2)))) =ln((x/( (√2)))+(√(1+(x^2 /2)))) ⇒ ∫F(u)du =a ln((x/2)+(√(1+(x^2 /2))))+b ln∣(x/( (√2)))+(√(1+(x^2 /2)))−α∣ +c ln∣(x/( (√2)))+(√(1+(x^2 /2)))+α∣ +c so the value of I is known.
$${I}=\int\:\frac{{x}+\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{{x}+\mathrm{1}−\sqrt{\mathrm{2}+{x}^{\mathrm{2}} }}{dx}\:{changement}\:{x}=\sqrt{\mathrm{2}}{sh}\left({t}\right)\:{give} \\ $$$${I}\:=\int\:\:\frac{\sqrt{\mathrm{2}}{sh}\left({t}\right)+\sqrt{\mathrm{2}}{ch}\left({t}\right)}{\:\sqrt{\mathrm{2}}{sh}\left({t}\right)+\mathrm{1}−\sqrt{\mathrm{2}}{cht}}\sqrt{\mathrm{2}}{cht}\:{dt} \\ $$$$=\:\int\:\frac{\left(\mathrm{2}{sh}\left({t}\right){ch}\left({t}\right)+{ch}^{\mathrm{2}} {t}\right)}{\:\sqrt{\mathrm{2}}{sh}\left({t}\right)−\sqrt{\mathrm{2}}{ch}\left({t}\right)+\mathrm{1}}{dt}\:=\int\:\:\frac{{sh}\left(\mathrm{2}{t}\right)+\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}{\:\sqrt{\mathrm{2}}{sh}\left({t}\right)−\sqrt{\mathrm{2}}{ch}\left({t}\right)+\mathrm{1}}{dt} \\ $$$$=\int\:\:\frac{\mathrm{2}{sh}\left(\mathrm{2}{t}\right)+{ch}\left(\mathrm{2}{t}\right)+\mathrm{1}}{\mathrm{2}\left(\sqrt{\mathrm{2}}{sh}\left({t}\right)−\sqrt{\mathrm{2}}{ch}\left({t}\right)\:+\mathrm{1}\right)}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\frac{\mathrm{2}\frac{{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} }{\mathrm{2}}+\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}\:+\mathrm{1}}{\:\sqrt{\mathrm{2}}\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}−\sqrt{\mathrm{2}}\frac{{e}^{{t}} \:+{e}^{−{t}} }{\mathrm{2}}+\mathrm{1}}{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\:\:\frac{\mathrm{2}{e}^{\mathrm{2}{t}} −\mathrm{2}{e}^{−\mathrm{2}{t}} \:+{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} \:+\mathrm{2}}{\:\sqrt{\mathrm{2}}{e}^{{t}} −\sqrt{\mathrm{2}}{e}^{−{t}} −\sqrt{\mathrm{2}}{e}^{{t}} −\sqrt{\mathrm{2}}{e}^{−{t}} \:+\mathrm{2}}{dt}\:\Rightarrow \\ $$$$\mathrm{2}{I}\:=\int\:\:\frac{\mathrm{3}{e}^{\mathrm{2}{t}} −{e}^{−\mathrm{2}{t}} \:+\mathrm{2}}{−\mathrm{2}\sqrt{\mathrm{2}}{e}^{−{t}} \:+\mathrm{2}}{dt}\:=_{{e}^{{t}} ={u}} \:\:\int\:\:\frac{\mathrm{3}{u}^{\mathrm{2}} −{u}^{−\mathrm{2}} \:+\mathrm{2}}{\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}{u}^{−\mathrm{1}} }\frac{{du}}{{u}} \\ $$$$=\int\:\:\frac{\mathrm{3}{u}^{\mathrm{2}} −{u}^{−\mathrm{2}} }{\mathrm{2}{u}−\mathrm{2}\sqrt{\mathrm{2}}}{du}\:=\int\:\:\:\frac{\mathrm{3}{u}^{\mathrm{4}} −\mathrm{1}}{\mathrm{2}{u}^{\mathrm{3}} −\mathrm{2}\sqrt{\mathrm{2}}{u}}{du}\:\Rightarrow \\ $$$$\mathrm{4}{I}\:=\:\int\:\:\frac{\mathrm{3}{u}^{\mathrm{4}} −\mathrm{1}}{{u}^{\mathrm{3}} −\sqrt{\mathrm{2}}{u}}{du}\:=\int\:\frac{\mathrm{3}{u}\left({u}^{\mathrm{3}} −\sqrt{\mathrm{2}}{u}\right)+\mathrm{3}\sqrt{\mathrm{2}}{u}^{\mathrm{2}} −\mathrm{1}}{{u}^{\mathrm{3}} −\sqrt{\mathrm{2}}{u}}{du} \\ $$$$=\mathrm{3}{u}\:+\int\:\:\frac{\mathrm{3}\sqrt{\mathrm{2}}{u}^{\mathrm{2}} −\mathrm{1}}{{u}\left({u}^{\mathrm{2}} −\sqrt{\mathrm{2}}\right)}{du}\:\:{let}\:{decompose}\:{F}\left({u}\right)=\frac{\mathrm{3}\sqrt{\mathrm{2}}{u}^{\mathrm{2}} −\mathrm{1}}{{u}\left({u}^{\mathrm{2}} −\sqrt{\mathrm{2}}\right)} \\ $$$$\Rightarrow{F}\left({u}\right)\:=\frac{\mathrm{3}\sqrt{\mathrm{2}}{u}^{\mathrm{2}} −\mathrm{1}}{{u}\left({u}−\alpha\right)\left({u}+\alpha\right)}\:\:{with}\:\alpha=^{\mathrm{4}} \sqrt{\mathrm{2}} \\ $$$${F}\left({u}\right)\:=\frac{{a}}{{u}}\:+\frac{{b}}{{u}−\alpha}\:+\frac{{c}}{{u}+\alpha} \\ $$$${a}\:=\frac{−\mathrm{1}}{−\alpha^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\alpha^{\mathrm{2}} }\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}} \\ $$$${b}\:=\frac{\mathrm{3}\sqrt{\mathrm{2}}\alpha^{\mathrm{2}} −\mathrm{1}}{\alpha\left(\mathrm{2}\alpha\right)}\:=\frac{\mathrm{3}\sqrt{\mathrm{2}}\alpha^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\alpha^{\mathrm{2}} }\:\:{and}\:{c}\:=\frac{\mathrm{3}\sqrt{\mathrm{2}}\alpha^{\mathrm{2}} −\mathrm{1}}{\left(−\alpha\right)\left(−\mathrm{2}\alpha\right)}\:=\frac{\mathrm{3}\sqrt{\mathrm{2}}\alpha^{\mathrm{2}} −\mathrm{1}}{\mathrm{2}\alpha^{\mathrm{2}} }\:{and} \\ $$$$\int\:{F}\left({u}\right){du}\:={aln}\mid{u}\mid+{bln}\mid{u}−\alpha\mid+{cln}\mid{u}+\alpha\mid\:{c} \\ $$$$={at}\:+{bln}\mid{e}^{{t}} −\alpha\mid\:+{cln}\mid{e}^{{t}} \:+\alpha\mid\:+{c}\:\:\:{but}\:{t}={argsh}\left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$={ln}\left(\frac{{x}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}\right)\:\Rightarrow \\ $$$$\int{F}\left({u}\right){du}\:={a}\:{ln}\left(\frac{{x}}{\mathrm{2}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}\right)+{b}\:{ln}\mid\frac{{x}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}−\alpha\mid \\ $$$$+{c}\:{ln}\mid\frac{{x}}{\:\sqrt{\mathrm{2}}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}+\alpha\mid\:+{c}\:\:{so}\:{the}\:{value}\:{of}\:{I}\:{is}\:{known}. \\ $$$$ \\ $$$$ \\ $$
Answered by MJS last updated on 06/Nov/19
2 substitutions in 1 step x=(√2)sinh s ∧ s=ln t ⇒ x=((t^2 −1)/(t(√2))) ⇔ t=((x+(√(x^2 +2)))/( (√2))) → dx=((√(2(x^2 +2)))/(x+(√(x^2 +2))))dt=((t^2 +1)/(t^2 (√2)))dt ∫((x+(√(x^2 +2)))/(x+1−(√(x^2 +2))))dx=∫((t^2 +1)/(t−(√2)))dt= =∫tdt+(√2)∫dt+3∫(dt/(t−(√2)))= =(1/2)t^2 +t(√2)+3ln (t−(√2))= =(((x+2)(x+(√(x^2 +2))))/2)+3ln (x−2+(√(x^2 +2))) +C
$$\mathrm{2}\:\mathrm{substitutions}\:\mathrm{in}\:\mathrm{1}\:\mathrm{step} \\ $$$${x}=\sqrt{\mathrm{2}}\mathrm{sinh}\:{s}\:\wedge\:{s}=\mathrm{ln}\:{t} \\ $$$$\Rightarrow \\ $$$${x}=\frac{{t}^{\mathrm{2}} −\mathrm{1}}{{t}\sqrt{\mathrm{2}}}\:\Leftrightarrow\:{t}=\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{\:\sqrt{\mathrm{2}}}\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{2}\left({x}^{\mathrm{2}} +\mathrm{2}\right)}}{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{dt}=\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}^{\mathrm{2}} \sqrt{\mathrm{2}}}{dt} \\ $$$$\int\frac{{x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{{x}+\mathrm{1}−\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}{dx}=\int\frac{{t}^{\mathrm{2}} +\mathrm{1}}{{t}−\sqrt{\mathrm{2}}}{dt}= \\ $$$$=\int{tdt}+\sqrt{\mathrm{2}}\int{dt}+\mathrm{3}\int\frac{{dt}}{{t}−\sqrt{\mathrm{2}}}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{t}^{\mathrm{2}} +{t}\sqrt{\mathrm{2}}+\mathrm{3ln}\:\left({t}−\sqrt{\mathrm{2}}\right)= \\ $$$$=\frac{\left({x}+\mathrm{2}\right)\left({x}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)}{\mathrm{2}}+\mathrm{3ln}\:\left({x}−\mathrm{2}+\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}\right)\:+{C} \\ $$
Commented by mathmax by abdo last updated on 07/Nov/19
thank you sir mjs.
$${thank}\:{you}\:{sir}\:{mjs}. \\ $$

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