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Question-81888

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Question Number 81888 by rajesh4661kumar@gmail.com last updated on 16/Feb/20
Commented by mathmax by abdo last updated on 16/Feb/20
let A =∫ (dx/((x^2 +1)(√(x^2 −1)))) we do the changement x =cht ⇒ A =∫ ((sht dt)/((ch^2 t +1)sht)) =∫ (dt/(1+((1+ch(2t))/2))) =∫ ((2dt)/(3+ch(2t))) =_(2t=u) ∫ (du/(3+ch(u))) =∫ (du/(3+((e^u +e^(−u) )/2))) =∫ ((2du)/(6 +e^u +e^(−u) )) =_(e^u =z) ∫ (2/(6+z +z^(−1) ))(dz/z) =∫ ((2dz)/(6z+z^2 +1)) =∫ ((2dz)/(z^2 +6z +1)) z^(2 ) +6z +1=0→Δ^′ =9−1 =8 ⇒z_1 =−3+2(√2) and z_2 =−3−2(√2)⇒ A =∫ ((2dz)/((z−z_1 )(z−z_2 ))) =(2/(z_1 −z_2 )) ∫((1/(z−z_1 ))−(1/(z−z_2 )))dz =(2/(4(√2)))ln∣((z−z_1 )/(z−z_2 ))∣ +C =(1/(2(√2)))ln∣((z+3−2(√2))/(z+3+2(√2)))∣ +C =(1/(2(√2)))ln∣((e^u +3−2(√2))/(e^u +3+2(√2)))∣ +C =(1/(2(√2)))ln∣((e^(2t) +3−2(√2))/(e^(2t) +3 +2(√2)))∣ +C we have t =argch(x) =ln(x+(√(x^2 −1))) ⇒2t =ln((x+(√(x^2 −1)))^2 ) ⇒e^(2t) =(x+(√(x^2 −1)))^2 =x^2 +2x(√(x^2 −1)) +x^2 −1 =2x^2 −1 +2x(√(x^2 −1)) ⇒ A =(1/(2(√2)))ln∣((2x^2 +2x(√(x^2 −1))+2−2(√2))/(2x^2 +2x(√(x^2 −1))+2+2(√3)))∣ +C =(1/(2(√2)))ln∣((x^2 +x(√(x^2 −1))+1−(√2))/(x^2 +x(√(x^2 −1)) +1+(√2)))∣ +C
$${let}\:{A}\:=\int\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}}\:\:{we}\:{do}\:{the}\:{changement}\:{x}\:={cht}\:\Rightarrow \\ $$$${A}\:=\int\:\:\:\frac{{sht}\:{dt}}{\left({ch}^{\mathrm{2}} {t}\:+\mathrm{1}\right){sht}}\:=\int\:\:\frac{{dt}}{\mathrm{1}+\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}\:=\int\:\:\frac{\mathrm{2}{dt}}{\mathrm{3}+{ch}\left(\mathrm{2}{t}\right)} \\ $$$$=_{\mathrm{2}{t}={u}} \:\:\int\:\:\frac{{du}}{\mathrm{3}+{ch}\left({u}\right)}\:=\int\:\:\:\frac{{du}}{\mathrm{3}+\frac{{e}^{{u}} \:+{e}^{−{u}} }{\mathrm{2}}}\:=\int\:\:\frac{\mathrm{2}{du}}{\mathrm{6}\:+{e}^{{u}} \:+{e}^{−{u}} } \\ $$$$=_{{e}^{{u}} ={z}} \:\:\:\int\:\:\frac{\mathrm{2}}{\mathrm{6}+{z}\:+{z}^{−\mathrm{1}} }\frac{{dz}}{{z}}\:=\int\:\:\frac{\mathrm{2}{dz}}{\mathrm{6}{z}+{z}^{\mathrm{2}} \:+\mathrm{1}}\:=\int\:\:\frac{\mathrm{2}{dz}}{{z}^{\mathrm{2}} \:+\mathrm{6}{z}\:+\mathrm{1}} \\ $$$${z}^{\mathrm{2}\:} \:+\mathrm{6}{z}\:+\mathrm{1}=\mathrm{0}\rightarrow\Delta^{'} =\mathrm{9}−\mathrm{1}\:=\mathrm{8}\:\Rightarrow{z}_{\mathrm{1}} =−\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}\:{and}\:{z}_{\mathrm{2}} =−\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}\Rightarrow \\ $$$${A}\:=\int\:\:\frac{\mathrm{2}{dz}}{\left({z}−{z}_{\mathrm{1}} \right)\left({z}−{z}_{\mathrm{2}} \right)}\:=\frac{\mathrm{2}}{{z}_{\mathrm{1}} −{z}_{\mathrm{2}} }\:\int\left(\frac{\mathrm{1}}{{z}−{z}_{\mathrm{1}} }−\frac{\mathrm{1}}{{z}−{z}_{\mathrm{2}} }\right){dz} \\ $$$$=\frac{\mathrm{2}}{\mathrm{4}\sqrt{\mathrm{2}}}{ln}\mid\frac{{z}−{z}_{\mathrm{1}} }{{z}−{z}_{\mathrm{2}} }\mid\:+{C}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{{z}+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{{z}+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{{e}^{{u}} \:\:+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{{e}^{{u}} \:+\mathrm{3}+\mathrm{2}\sqrt{\mathrm{2}}}\mid\:+{C}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{{e}^{\mathrm{2}{t}} \:+\mathrm{3}−\mathrm{2}\sqrt{\mathrm{2}}}{{e}^{\mathrm{2}{t}} +\mathrm{3}\:+\mathrm{2}\sqrt{\mathrm{2}}}\mid\:+{C} \\ $$$${we}\:{have}\:{t}\:={argch}\left({x}\right)\:={ln}\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)\:\Rightarrow\mathrm{2}{t}\:={ln}\left(\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} \right) \\ $$$$\Rightarrow{e}^{\mathrm{2}{t}} =\left({x}+\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\right)^{\mathrm{2}} ={x}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:+{x}^{\mathrm{2}} −\mathrm{1} \\ $$$$=\mathrm{2}{x}^{\mathrm{2}} −\mathrm{1}\:+\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\mathrm{2}−\mathrm{2}\sqrt{\mathrm{2}}}{\mathrm{2}{x}^{\mathrm{2}} +\mathrm{2}{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\mathrm{2}+\mathrm{2}\sqrt{\mathrm{3}}}\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}\sqrt{\mathrm{2}}}{ln}\mid\frac{{x}^{\mathrm{2}} \:+{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}+\mathrm{1}−\sqrt{\mathrm{2}}}{{x}^{\mathrm{2}} \:+{x}\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:+\mathrm{1}+\sqrt{\mathrm{2}}}\mid\:+{C} \\ $$
Answered by TANMAY PANACEA last updated on 16/Feb/20
t^2 =((x^2 −1)/(x^2 +1))→x^2 t^2 +t^2 =x^2 −1 x^2 (t^2 −1)=−t^2 −1 x^2 =((1+t^2 )/(1−t^2 ))→x=(√((1+t^2 )/(1−t^2 ))) dx=(1/2)(((1+t^2 )/(1−t^2 )))^((−1)/2) ×(((1−t^2 ).2t−(1+t^2 )(−2t))/((1−t^2 )^2 ))dt dx=(1/2)×(√((1−t^2 )/(1+t^2 ))) ×((2t−2t^3 +2t+2t^3 )/((1−t^2 )^2 ))dt ∫((2tdt)/( (√(1+t^2 )) ×(1−t^2 )^(3/2) ))×(1/((((1+t^2 )/(1−t^2 ))+1)((√(((1+t^2 )/(1−t^2 ))−1)) )) ∫((2tdt)/( (√(1+t^2 )) (1−t^2 )^(3/2) ×(2/(1−t^2 ))×(√((1+t^2 −1+t^2 )/(1−t^2 ))))) ∫((2tdt)/( (√(1+t^2 ))))×(1/( (√2) t)) (√2) ∫(dt/( (√(1+t^2 ))))=(√2) ln(t+(√(1+t^2 )) )+c now (√2) ln((√(((x^2 −1)/(x^2 +1)) )) +(√(1+((x^2 −1)/(x^2 +1)))) )+c (√2) ln((((√(x^2 −1)) +x(√2))/( (√(x^2 +1)))))+c pls check...
$${t}^{\mathrm{2}} =\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\rightarrow{x}^{\mathrm{2}} {t}^{\mathrm{2}} +{t}^{\mathrm{2}} ={x}^{\mathrm{2}} −\mathrm{1} \\ $$$${x}^{\mathrm{2}} \left({t}^{\mathrm{2}} −\mathrm{1}\right)=−{t}^{\mathrm{2}} −\mathrm{1} \\ $$$${x}^{\mathrm{2}} =\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\rightarrow{x}=\sqrt{\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }}\: \\ $$$${dx}=\frac{\mathrm{1}}{\mathrm{2}}\left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }\right)^{\frac{−\mathrm{1}}{\mathrm{2}}} ×\frac{\left(\mathrm{1}−{t}^{\mathrm{2}} \right).\mathrm{2}{t}−\left(\mathrm{1}+{t}^{\mathrm{2}} \right)\left(−\mathrm{2}{t}\right)}{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$${dx}=\frac{\mathrm{1}}{\mathrm{2}}×\sqrt{\frac{\mathrm{1}−{t}^{\mathrm{2}} }{\mathrm{1}+{t}^{\mathrm{2}} }}\:×\frac{\mathrm{2}{t}−\mathrm{2}{t}^{\mathrm{3}} +\mathrm{2}{t}+\mathrm{2}{t}^{\mathrm{3}} }{\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\mathrm{2}} }{dt} \\ $$$$\int\frac{\mathrm{2}{tdt}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:×\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} }×\frac{\mathrm{1}}{\left(\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }+\mathrm{1}\right)\left(\sqrt{\frac{\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }−\mathrm{1}}\:\right.} \\ $$$$\int\frac{\mathrm{2}{tdt}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:\left(\mathrm{1}−{t}^{\mathrm{2}} \right)^{\frac{\mathrm{3}}{\mathrm{2}}} ×\frac{\mathrm{2}}{\mathrm{1}−{t}^{\mathrm{2}} }×\sqrt{\frac{\mathrm{1}+{t}^{\mathrm{2}} −\mathrm{1}+{t}^{\mathrm{2}} }{\mathrm{1}−{t}^{\mathrm{2}} }}} \\ $$$$\int\frac{\mathrm{2}{tdt}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}×\frac{\mathrm{1}}{\:\sqrt{\mathrm{2}}\:{t}} \\ $$$$\sqrt{\mathrm{2}}\:\int\frac{{dt}}{\:\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }}=\sqrt{\mathrm{2}}\:{ln}\left({t}+\sqrt{\mathrm{1}+{t}^{\mathrm{2}} }\:\right)+{c} \\ $$$${now}\:\sqrt{\mathrm{2}}\:{ln}\left(\sqrt{\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\:}\:+\sqrt{\mathrm{1}+\frac{{x}^{\mathrm{2}} −\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}}\:\right)+{c} \\ $$$$\sqrt{\mathrm{2}}\:{ln}\left(\frac{\sqrt{{x}^{\mathrm{2}} −\mathrm{1}}\:+{x}\sqrt{\mathrm{2}}}{\:\sqrt{{x}^{\mathrm{2}} +\mathrm{1}}}\right)+{c} \\ $$$$\boldsymbol{{pls}}\:\boldsymbol{{check}}… \\ $$$$ \\ $$

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