dx-x-1-x-2-2-

Question Number 44069 by ajfour last updated on 21/Sep/18

$$\int\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}\:=\:? \\ $$

Commented by maxmathsup by imad last updated on 21/Sep/18

let I = ∫ (dx/((x+1)(√(x^2 +2)))) changement x=(√2)sh(t)give I = ∫ (((√2)ch(t)dt)/((1+(√2)sh(t))(√2)ch(t))) = ∫ (dt/(1+(√2)((e^t −e^(−t) )/2))) = ∫ ((2dt)/(2 +(√2)e^t −(√2)e^(−t) )) =_(e^t =u) ∫ (2/(2 +(√2)u−(√2)u^(−1) )) (du/u) = ∫ ((2du)/(2u +(√2)u^2 −(√2))) = ∫ ((2du)/( (√2)u^2 +2u −(√2))) =(√2) ∫ (du/(u^2 +(√2)u −1)) let decompose F(u) = (1/(u^2 +(√2)u−1)) Δ =2−4(−1) =6 ⇒u_1 =((−(√2)+(√6))/2) and u_2 =((−(√2) −(√6))/2) F(u) = (a/(u−u_1 )) +(b/(u−u_2 )) =(1/((u−u_1 )(u−u_2 ))) ⇒ a =(1/(u_1 −u_2 )) = (1/( (√6))) and b =(1/(u_2 −u_1 )) =−(1/( (√6))) ⇒ F(u) =(1/( (√6))){ (1/(u−u_1 )) −(1/(u−u_2 ))} ⇒I =(1/( (√6)))∫ ((1/(u−u_1 )) −(1/(u−u_2 )))du =(1/( (√6)))ln∣((u−u_1 )/(u−u_2 ))∣ +c =(1/( (√6)))ln∣ ((e^t −u_1 )/(e^t −u_2 ))∣ +c but t =argsh((x/( (√2)))) =ln((x/( (√2))) +(√(1+(x^2 /2)))) ⇒e^t =(x/2) +(√(1+(x^2 /2)))⇒ I =(1/( (√6)))ln∣ (((x/2)+(√(1+(x^2 /2)))+(((√2)−(√6))/2))/((x/2)+(√(1+(x^2 /2)))+(((√2)+(√6))/2)))∣ +c.

$${let}\:{I}\:=\:\int\:\:\:\frac{{dx}}{\left({x}+\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +\mathrm{2}}}\:\:{changement}\:{x}=\sqrt{\mathrm{2}}{sh}\left({t}\right){give} \\ $$$${I}\:=\:\int\:\:\:\:\:\:\frac{\sqrt{\mathrm{2}}{ch}\left({t}\right){dt}}{\left(\mathrm{1}+\sqrt{\mathrm{2}}{sh}\left({t}\right)\right)\sqrt{\mathrm{2}}{ch}\left({t}\right)}\:=\:\int\:\:\frac{{dt}}{\mathrm{1}+\sqrt{\mathrm{2}}\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}} \\ $$$$=\:\int\:\:\:\frac{\mathrm{2}{dt}}{\mathrm{2}\:+\sqrt{\mathrm{2}}{e}^{{t}} −\sqrt{\mathrm{2}}{e}^{−{t}} }\:=_{{e}^{{t}} ={u}} \:\:\:\:\:\int\:\:\:\:\:\:\frac{\mathrm{2}}{\mathrm{2}\:+\sqrt{\mathrm{2}}{u}−\sqrt{\mathrm{2}}{u}^{−\mathrm{1}} }\:\frac{{du}}{{u}} \\ $$$$=\:\int\:\:\:\:\frac{\mathrm{2}{du}}{\mathrm{2}{u}\:+\sqrt{\mathrm{2}}{u}^{\mathrm{2}} −\sqrt{\mathrm{2}}}\:=\:\int\:\:\:\frac{\mathrm{2}{du}}{\:\sqrt{\mathrm{2}}{u}^{\mathrm{2}} \:+\mathrm{2}{u}\:−\sqrt{\mathrm{2}}}\:=\sqrt{\mathrm{2}}\:\int\:\:\:\:\frac{{du}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}\:−\mathrm{1}} \\ $$$${let}\:{decompose}\:{F}\left({u}\right)\:=\:\frac{\mathrm{1}}{{u}^{\mathrm{2}} \:+\sqrt{\mathrm{2}}{u}−\mathrm{1}} \\ $$$$\Delta\:=\mathrm{2}−\mathrm{4}\left(−\mathrm{1}\right)\:=\mathrm{6}\:\:\Rightarrow{u}_{\mathrm{1}} =\frac{−\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}{\mathrm{2}}\:\:{and}\:{u}_{\mathrm{2}} =\frac{−\sqrt{\mathrm{2}}\:−\sqrt{\mathrm{6}}}{\mathrm{2}} \\ $$$${F}\left({u}\right)\:=\:\frac{{a}}{{u}−{u}_{\mathrm{1}} }\:+\frac{{b}}{{u}−{u}_{\mathrm{2}} }\:=\frac{\mathrm{1}}{\left({u}−{u}_{\mathrm{1}} \right)\left({u}−{u}_{\mathrm{2}} \right)}\:\Rightarrow \\ $$$${a}\:=\frac{\mathrm{1}}{{u}_{\mathrm{1}} −{u}_{\mathrm{2}} }\:=\:\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\:\:{and}\:{b}\:=\frac{\mathrm{1}}{{u}_{\mathrm{2}} −{u}_{\mathrm{1}} }\:=−\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\:\Rightarrow \\ $$$${F}\left({u}\right)\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\left\{\:\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right\}\:\Rightarrow{I}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}\int\:\left(\frac{\mathrm{1}}{{u}−{u}_{\mathrm{1}} }\:−\frac{\mathrm{1}}{{u}−{u}_{\mathrm{2}} }\right){du} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}{ln}\mid\frac{{u}−{u}_{\mathrm{1}} }{{u}−{u}_{\mathrm{2}} }\mid\:+{c}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}{ln}\mid\:\frac{{e}^{{t}} \:−{u}_{\mathrm{1}} }{{e}^{{t}} −{u}_{\mathrm{2}} }\mid\:+{c}\:{but}\:{t}\:={argsh}\left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\right) \\ $$$$={ln}\left(\frac{{x}}{\:\sqrt{\mathrm{2}}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}\right)\:\Rightarrow{e}^{{t}} \:=\frac{{x}}{\mathrm{2}}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}\Rightarrow \\ $$$${I}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{6}}}{ln}\mid\:\frac{\frac{{x}}{\mathrm{2}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}+\frac{\sqrt{\mathrm{2}}−\sqrt{\mathrm{6}}}{\mathrm{2}}}{\frac{{x}}{\mathrm{2}}+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} }{\mathrm{2}}}+\frac{\sqrt{\mathrm{2}}+\sqrt{\mathrm{6}}}{\mathrm{2}}}\mid\:+{c}. \\ $$

Answered by tanmay.chaudhury50@gmail.com last updated on 21/Sep/18

t=(1/(x+1)) x+1=(1/t) dx=−(1/t^2 )dt ∫((−dt)/(t^2 ×(1/t)×(√(((1/t)−1)^2 +2)))) ∫((−dt)/(t×(√((1/t^2 )−(2/t)+1+2)))) ∫((−dt)/(t×(√((1−2t+3t^2 )/t^2 )))) ∫((−dt)/( (√(3(t^2 −(2/3)t+(1/3)))))) =((−1)/( (√3)))∫(dt/( (√(t^2 −2.t.(1/3)+(1/9)+(1/3)−(1/9))))) =((−1)/3)∫(dt/( (√((t−(1/3))^2 +((((√2) )/3))^2 )))) =((−1)/3)ln∣(t−(1/3))+(√((t−(1/3))^2 +((((√2) )/3))^2 ∣)) =((−1)/3)ln∣((1/(x+1))−(1/3))+(√(((1/(x+1))−(1/3))^2 +((2/9))))

$${t}=\frac{\mathrm{1}}{{x}+\mathrm{1}}\:\:{x}+\mathrm{1}=\frac{\mathrm{1}}{{t}}\:\:\:{dx}=−\frac{\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$$\int\frac{−{dt}}{{t}^{\mathrm{2}} ×\frac{\mathrm{1}}{{t}}×\sqrt{\left(\frac{\mathrm{1}}{{t}}−\mathrm{1}\right)^{\mathrm{2}} +\mathrm{2}}} \\ $$$$\int\frac{−{dt}}{{t}×\sqrt{\frac{\mathrm{1}}{{t}^{\mathrm{2}} }−\frac{\mathrm{2}}{{t}}+\mathrm{1}+\mathrm{2}}} \\ $$$$\int\frac{−{dt}}{{t}×\sqrt{\frac{\mathrm{1}−\mathrm{2}{t}+\mathrm{3}{t}^{\mathrm{2}} }{{t}^{\mathrm{2}} }}} \\ $$$$\int\frac{−{dt}}{\:\sqrt{\mathrm{3}\left({t}^{\mathrm{2}} −\frac{\mathrm{2}}{\mathrm{3}}{t}+\frac{\mathrm{1}}{\mathrm{3}}\right)}} \\ $$$$=\frac{−\mathrm{1}}{\:\sqrt{\mathrm{3}}}\int\frac{{dt}}{\:\sqrt{{t}^{\mathrm{2}} −\mathrm{2}.{t}.\frac{\mathrm{1}}{\mathrm{3}}+\frac{\mathrm{1}}{\mathrm{9}}+\frac{\mathrm{1}}{\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{9}}}} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{3}}\int\frac{{dt}}{\:\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{2}}\:}{\mathrm{3}}\right)^{\mathrm{2}} }} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{3}}{ln}\mid\left({t}−\frac{\mathrm{1}}{\mathrm{3}}\right)+\sqrt{\left({t}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{2}}\:}{\mathrm{3}}\right)^{\mathrm{2}} \:\mid} \\ $$$$=\frac{−\mathrm{1}}{\mathrm{3}}{ln}\mid\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\right)+\sqrt{\left(\frac{\mathrm{1}}{{x}+\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{3}}\right)^{\mathrm{2}} +\left(\frac{\mathrm{2}}{\mathrm{9}}\right)}\: \\ $$$$ \\ $$

Commented by ajfour last updated on 21/Sep/18