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Question-120092

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Question Number 120092 by danielasebhofoh last updated on 29/Oct/20
Commented by Lordose last updated on 29/Oct/20
SMALL EINSTEIN..?��
Answered by Lordose last updated on 29/Oct/20
Ω = ∫((2x^2 +5)/(x^3 (√(x^4 +x^2 +1))))dx set u= x^2 ⇒ du=2xdx Ω = (1/2)∫((2u+5)/(u^2 (√(u^2 +u+1))))du Ω = (1/2)∫(((1/u^2 )(2u+5))/( (√((u+(1/2))^2 + (3/4)))))du set u + (1/2) = y ⇒ du=dy Ω = (1/2)∫(((2/((y−(1/2)))) + (5/((y−(1/2))^2 )))/( (√(y^2 + (3/4)))))dy set y = ((√3)/2)tanθ ⇒ dy = ((√3)/2)sec^2 θdθ Ω = (1/2)∫ ((((√3)/2)secθ((2/( (1/2)((√3)tanθ−1))) + (5/((1/4)((√3)tanθ−1)^2 ))))/((√3)/2))dθ Ω = (1/2)∫(((4secθ)/( (√3)tanθ−1)) + ((20secθ)/( ((√3)tanθ−1)^2 )))dθ It′s easy from here
$$\Omega\:=\:\int\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{5}}{\mathrm{x}^{\mathrm{3}} \sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\mathrm{dx} \\ $$$$\mathrm{set}\:\mathrm{u}=\:\mathrm{x}^{\mathrm{2}} \:\Rightarrow\:\mathrm{du}=\mathrm{2xdx} \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2u}+\mathrm{5}}{\mathrm{u}^{\mathrm{2}} \sqrt{\mathrm{u}^{\mathrm{2}} +\mathrm{u}+\mathrm{1}}}\mathrm{du} \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\frac{\mathrm{1}}{\mathrm{u}^{\mathrm{2}} }\left(\mathrm{2u}+\mathrm{5}\right)}{\:\sqrt{\left(\mathrm{u}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\:\frac{\mathrm{3}}{\mathrm{4}}}}\mathrm{du} \\ $$$$\mathrm{set}\:\mathrm{u}\:+\:\frac{\mathrm{1}}{\mathrm{2}}\:=\:\mathrm{y}\:\Rightarrow\:\mathrm{du}=\mathrm{dy} \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\frac{\mathrm{2}}{\left(\mathrm{y}−\frac{\mathrm{1}}{\mathrm{2}}\right)}\:+\:\frac{\mathrm{5}}{\left(\mathrm{y}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} }}{\:\sqrt{\mathrm{y}^{\mathrm{2}} \:+\:\frac{\mathrm{3}}{\mathrm{4}}}}\mathrm{dy} \\ $$$$\mathrm{set}\:\mathrm{y}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{tan}\theta\:\Rightarrow\:\mathrm{dy}\:=\:\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sec}^{\mathrm{2}} \theta\mathrm{d}\theta \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{sec}\theta\left(\frac{\mathrm{2}}{\:\frac{\mathrm{1}}{\mathrm{2}}\left(\sqrt{\mathrm{3}}\mathrm{tan}\theta−\mathrm{1}\right)}\:+\:\frac{\mathrm{5}}{\frac{\mathrm{1}}{\mathrm{4}}\left(\sqrt{\mathrm{3}}\mathrm{tan}\theta−\mathrm{1}\right)^{\mathrm{2}} }\right)}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}}\mathrm{d}\theta \\ $$$$\Omega\:=\:\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{\mathrm{4sec}\theta}{\:\sqrt{\mathrm{3}}\mathrm{tan}\theta−\mathrm{1}}\:+\:\frac{\mathrm{20sec}\theta}{\:\left(\sqrt{\mathrm{3}}\mathrm{tan}\theta−\mathrm{1}\right)^{\mathrm{2}} }\right)\mathrm{d}\theta \\ $$$$\mathrm{It}'\mathrm{s}\:\mathrm{easy}\:\mathrm{from}\:\mathrm{here} \\ $$
Answered by mathmax by abdo last updated on 29/Oct/20
A =∫ ((2x^2 +5)/(x^3 (√(x^4 +x^2 +1))))dx changement x^2 =t give A =∫ ((2t+5)/(t(√t)(√(t^2 +t+1))))(dt/(2(√t))) =∫ ((2t+5)/(2t^2 (√(t^2 +t+1))))dt =(1/2)∫ ((2t+5)/(t^2 (√((t+(1/2))^2 +(3/4))))) =_(t+(1/2)=((√3)/2)shu) (1/2)∫ ((2(((√3)/2)shu−(1/2))+5)/((((√3)/2)shu−(1/2))^2 (√(3/4))chu))((√3)/2)chudu =2∫ (((√3)shu+4)/(((√3)shu−1)^2 ))du =2 ∫ (((√3)((e^u −e^(−u) )/2)+4)/(((√3)((e^u −e^(−u) )/2)−1)^2 ))du =4∫ (((√3)e^u −(√3)e^(−u) +8)/(((√3)(e^u −e^(−u) )−2)^2 ))du =4 ∫ (((√3)e^u −(√3)e^(−u) +8)/(3(e^u −e^(−u) )^2 −4(√3)(e^u −e^(−u) )+4))du =4∫ (((√3)e^u −(√3)e^(−u) +8)/(3(e^(2u) −2+e^(−2u) )−4(√3)e^u +4(√3)e^(−u) +4))du =4∫ (((√3)e^u −(√3)e^(−u) +8)/(3e^(2u) −2 +3e^(−2u) −4(√3)e^u +4(√3)e^(−u) ))du =4∫ (((√3)e^(3u) −(√3)e^u +8e^(2u) )/(3e^(4u) +3−4(√3)e^(3u) +4(√3)u−2e^(2u) ))du =_(e^u =z) 4∫ (((√3)z^3 +8z^2 −(√3)z)/(3z^4 −4(√3)z^3 −2z^2 +4(√3)z +3))(dz/z) =4∫ (((√3)z^2 +8z −(√3))/(3z^4 −4(√3)z^3 −2z^2 +4(√3)z +3))dx rest decoposition...be continued .......
$$\mathrm{A}\:=\int\:\:\frac{\mathrm{2x}^{\mathrm{2}} +\mathrm{5}}{\mathrm{x}^{\mathrm{3}} \sqrt{\mathrm{x}^{\mathrm{4}} +\mathrm{x}^{\mathrm{2}} +\mathrm{1}}}\mathrm{dx}\:\:\mathrm{changement}\:\mathrm{x}^{\mathrm{2}} =\mathrm{t}\:\mathrm{give} \\ $$$$\mathrm{A}\:=\int\:\:\:\frac{\mathrm{2t}+\mathrm{5}}{\mathrm{t}\sqrt{\mathrm{t}}\sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{t}+\mathrm{1}}}\frac{\mathrm{dt}}{\mathrm{2}\sqrt{\mathrm{t}}}\:=\int\:\:\:\frac{\mathrm{2t}+\mathrm{5}}{\mathrm{2t}^{\mathrm{2}} \sqrt{\mathrm{t}^{\mathrm{2}} +\mathrm{t}+\mathrm{1}}}\mathrm{dt} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\:\:\frac{\mathrm{2t}+\mathrm{5}}{\mathrm{t}^{\mathrm{2}} \sqrt{\left(\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}}\:=_{\mathrm{t}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{shu}} \frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{shu}−\frac{\mathrm{1}}{\mathrm{2}}\right)+\mathrm{5}}{\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{shu}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \sqrt{\frac{\mathrm{3}}{\mathrm{4}}}\mathrm{chu}}\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\mathrm{chudu} \\ $$$$=\mathrm{2}\int\:\:\frac{\sqrt{\mathrm{3}}\mathrm{shu}+\mathrm{4}}{\left(\sqrt{\mathrm{3}}\mathrm{shu}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{du}\:=\mathrm{2}\:\int\:\:\frac{\sqrt{\mathrm{3}}\frac{\mathrm{e}^{\mathrm{u}} −\mathrm{e}^{−\mathrm{u}} }{\mathrm{2}}+\mathrm{4}}{\left(\sqrt{\mathrm{3}}\frac{\mathrm{e}^{\mathrm{u}} −\mathrm{e}^{−\mathrm{u}} }{\mathrm{2}}−\mathrm{1}\right)^{\mathrm{2}} }\mathrm{du} \\ $$$$=\mathrm{4}\int\:\:\frac{\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{u}} −\sqrt{\mathrm{3}}\mathrm{e}^{−\mathrm{u}} +\mathrm{8}}{\left(\sqrt{\mathrm{3}}\left(\mathrm{e}^{\mathrm{u}} −\mathrm{e}^{−\mathrm{u}} \right)−\mathrm{2}\right)^{\mathrm{2}} }\mathrm{du}\:=\mathrm{4}\:\int\:\:\frac{\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{u}} −\sqrt{\mathrm{3}}\mathrm{e}^{−\mathrm{u}} +\mathrm{8}}{\mathrm{3}\left(\mathrm{e}^{\mathrm{u}} −\mathrm{e}^{−\mathrm{u}} \right)^{\mathrm{2}} −\mathrm{4}\sqrt{\mathrm{3}}\left(\mathrm{e}^{\mathrm{u}} −\mathrm{e}^{−\mathrm{u}} \right)+\mathrm{4}}\mathrm{du} \\ $$$$=\mathrm{4}\int\:\:\frac{\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{u}} −\sqrt{\mathrm{3}}\mathrm{e}^{−\mathrm{u}} +\mathrm{8}}{\mathrm{3}\left(\mathrm{e}^{\mathrm{2u}} −\mathrm{2}+\mathrm{e}^{−\mathrm{2u}} \right)−\mathrm{4}\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{u}} +\mathrm{4}\sqrt{\mathrm{3}}\mathrm{e}^{−\mathrm{u}} +\mathrm{4}}\mathrm{du} \\ $$$$=\mathrm{4}\int\:\frac{\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{u}} −\sqrt{\mathrm{3}}\mathrm{e}^{−\mathrm{u}} +\mathrm{8}}{\mathrm{3e}^{\mathrm{2u}} −\mathrm{2}\:+\mathrm{3e}^{−\mathrm{2u}} −\mathrm{4}\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{u}} +\mathrm{4}\sqrt{\mathrm{3}}\mathrm{e}^{−\mathrm{u}} }\mathrm{du} \\ $$$$=\mathrm{4}\int\:\:\frac{\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{3u}} −\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{u}} +\mathrm{8e}^{\mathrm{2u}} }{\mathrm{3e}^{\mathrm{4u}} +\mathrm{3}−\mathrm{4}\sqrt{\mathrm{3}}\mathrm{e}^{\mathrm{3u}} +\mathrm{4}\sqrt{\mathrm{3}}\mathrm{u}−\mathrm{2e}^{\mathrm{2u}} }\mathrm{du} \\ $$$$=_{\mathrm{e}^{\mathrm{u}} =\mathrm{z}} \:\:\:\:\mathrm{4}\int\:\:\:\frac{\sqrt{\mathrm{3}}\mathrm{z}^{\mathrm{3}} +\mathrm{8z}^{\mathrm{2}} −\sqrt{\mathrm{3}}\mathrm{z}}{\mathrm{3z}^{\mathrm{4}} −\mathrm{4}\sqrt{\mathrm{3}}\mathrm{z}^{\mathrm{3}} −\mathrm{2z}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\mathrm{z}\:+\mathrm{3}}\frac{\mathrm{dz}}{\mathrm{z}} \\ $$$$=\mathrm{4}\int\:\frac{\sqrt{\mathrm{3}}\mathrm{z}^{\mathrm{2}} +\mathrm{8z}\:−\sqrt{\mathrm{3}}}{\mathrm{3z}^{\mathrm{4}} −\mathrm{4}\sqrt{\mathrm{3}}\mathrm{z}^{\mathrm{3}} −\mathrm{2z}^{\mathrm{2}} +\mathrm{4}\sqrt{\mathrm{3}}\mathrm{z}\:+\mathrm{3}}\mathrm{dx}\:\:\mathrm{rest}\:\mathrm{decoposition}…\mathrm{be}\:\mathrm{continued} \\ $$$$……. \\ $$
Answered by TANMAY PANACEA last updated on 29/Oct/20
∫(((2/x^3 )+(5/x^5 ))/( (√(1+(1/x^2 )+(1/x^4 )))))dx t^2 =1+x^(−2) +x^(−4) 2tdt=(((−2)/x^3 )+((−4)/x^5 ))dx −2tdt=((2/x^3 )+(4/x^5 ))dx ∫(((2/x^3 )+(4/x^5 ))/( (√(1+(1/x^2 )+(1/x^4 )))))dx+∫((1/x^5 )/( (√(1+(1/x^2 )+(1/x^4 )))))dx I=I_1 +I_2 I_1 =∫((−2tdt)/t)=−2t=−2((√(1+(1/x^2 )+(1/x^4 ))) )+c_1 I_2 =∫(dx/(x^5 (√((x^4 +x^2 +1)/x^4 )))) =∫(dx/(x^3 (√(x^4 +x^2 +1))))→k=x^2 dk=2xdx (1/2)∫((2xdx)/(x^4 (√(x^4 +x^2 +1)))) (1/2)∫(dk/(k^2 (√(k^2 +k+1)))) Wait p=(1/k)→dp=((−dk)/k^2 ) (1/2)∫((−dp)/( (√((1/p^2 )+(1/p)+1)))) =(((−1)/2))∫((pdp)/( (√(p^2 +p+1)))) =(((−1)/4))∫((2p+1−1)/( (√(p^2 +p+1))))dp =(((−1)/4))∫((d(p^2 +p+1))/( (√(p^2 +p+1))))+(1/4)∫(dp/( (√(p^2 +2p×(1/2)+(1/4)+(3/4))))) =(((−1)/4))×((√(p^2 +p+1))/(1/2))+∫(dp/( (√((p+(1/2))^2 +(((√3)/2))^2 )))) =(((−1)/2))(√(p^2 +p+1)) +ln{(p+(1/2))+(√((p+(1/2))^2 +(((√3)/2))^2 )) +c_2 p=(1/k) and k=x^2 so replace p by=(1/x^2 ) I_2 =(((−1)/2))(√((1/x^4 )+(1/x^2 )+1)) +ln{((1/x^2 )+(1/2))+(√(((1/x^2 )+(1/2))^2 +(((√3)/2))^2 )) +c_2 now please add I_1 and I_(2 ) →I=I_1 +I_2
$$\int\frac{\frac{\mathrm{2}}{{x}^{\mathrm{3}} }+\frac{\mathrm{5}}{{x}^{\mathrm{5}} }}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}{dx} \\ $$$${t}^{\mathrm{2}} =\mathrm{1}+{x}^{−\mathrm{2}} +{x}^{−\mathrm{4}} \\ $$$$\mathrm{2}{tdt}=\left(\frac{−\mathrm{2}}{{x}^{\mathrm{3}} }+\frac{−\mathrm{4}}{{x}^{\mathrm{5}} }\right){dx} \\ $$$$−\mathrm{2}{tdt}=\left(\frac{\mathrm{2}}{{x}^{\mathrm{3}} }+\frac{\mathrm{4}}{{x}^{\mathrm{5}} }\right){dx} \\ $$$$\int\frac{\frac{\mathrm{2}}{{x}^{\mathrm{3}} }+\frac{\mathrm{4}}{{x}^{\mathrm{5}} }}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}{dx}+\int\frac{\frac{\mathrm{1}}{{x}^{\mathrm{5}} }}{\:\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}}{dx} \\ $$$$\boldsymbol{{I}}=\boldsymbol{{I}}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$$${I}_{\mathrm{1}} =\int\frac{−\mathrm{2}{tdt}}{{t}}=−\mathrm{2}{t}=−\mathrm{2}\left(\sqrt{\mathrm{1}+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{{x}^{\mathrm{4}} }}\:\right)+{c}_{\mathrm{1}} \\ $$$${I}_{\mathrm{2}} =\int\frac{{dx}}{{x}^{\mathrm{5}} \sqrt{\frac{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}{{x}^{\mathrm{4}} }}} \\ $$$$=\int\frac{{dx}}{{x}^{\mathrm{3}} \sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}}\rightarrow{k}={x}^{\mathrm{2}} \:\:\:{dk}=\mathrm{2}{xdx} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{xdx}}{{x}^{\mathrm{4}} \sqrt{{x}^{\mathrm{4}} +{x}^{\mathrm{2}} +\mathrm{1}}} \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{{dk}}{{k}^{\mathrm{2}} \sqrt{{k}^{\mathrm{2}} +{k}+\mathrm{1}}} \\ $$$${Wait} \\ $$$${p}=\frac{\mathrm{1}}{{k}}\rightarrow{dp}=\frac{−{dk}}{{k}^{\mathrm{2}} } \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\int\frac{−{dp}}{\:\sqrt{\frac{\mathrm{1}}{{p}^{\mathrm{2}} }+\frac{\mathrm{1}}{{p}}+\mathrm{1}}} \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\int\frac{{pdp}}{\:\sqrt{{p}^{\mathrm{2}} +{p}+\mathrm{1}}} \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{4}}\right)\int\frac{\mathrm{2}{p}+\mathrm{1}−\mathrm{1}}{\:\sqrt{{p}^{\mathrm{2}} +{p}+\mathrm{1}}}{dp} \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{4}}\right)\int\frac{{d}\left({p}^{\mathrm{2}} +{p}+\mathrm{1}\right)}{\:\sqrt{{p}^{\mathrm{2}} +{p}+\mathrm{1}}}+\frac{\mathrm{1}}{\mathrm{4}}\int\frac{{dp}}{\:\sqrt{{p}^{\mathrm{2}} +\mathrm{2}{p}×\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{4}}\right)×\frac{\sqrt{{p}^{\mathrm{2}} +{p}+\mathrm{1}}}{\frac{\mathrm{1}}{\mathrm{2}}}+\int\frac{{dp}}{\:\sqrt{\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }} \\ $$$$=\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\sqrt{{p}^{\mathrm{2}} +{p}+\mathrm{1}}\:+{ln}\left\{\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\left({p}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+{c}_{\mathrm{2}} \right. \\ $$$$\boldsymbol{{p}}=\frac{\mathrm{1}}{\boldsymbol{{k}}}\:\:\boldsymbol{{and}}\:\boldsymbol{{k}}=\boldsymbol{{x}}^{\mathrm{2}} \:\:\boldsymbol{{so}}\:\boldsymbol{{replace}}\:\boldsymbol{{p}}\:\boldsymbol{{by}}=\frac{\mathrm{1}}{\boldsymbol{{x}}^{\mathrm{2}} } \\ $$$$\boldsymbol{{I}}_{\mathrm{2}} =\left(\frac{−\mathrm{1}}{\mathrm{2}}\right)\sqrt{\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\mathrm{1}}\:+{ln}\left\{\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\right)+\sqrt{\left(\frac{\mathrm{1}}{{x}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }\:+{c}_{\mathrm{2}} \right. \\ $$$${now}\:{please}\:{add}\:{I}_{\mathrm{1}} \:{and}\:{I}_{\mathrm{2}\:} \rightarrow{I}={I}_{\mathrm{1}} +{I}_{\mathrm{2}} \\ $$
Commented by peter frank last updated on 30/Oct/20
thank you both
$$\mathrm{thank}\:\mathrm{you}\:\mathrm{both} \\ $$

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