Menu Close

1-find-x-2-x-1-x-2-3-dx-2-calculate-0-1-x-2-x-1-x-2-3-dx-




Question Number 82442 by mathmax by abdo last updated on 21/Feb/20
1)find ∫ ((√(x^2 −x+1))/(x^2  +3))dx  2)calculate ∫_0 ^1  ((√(x^2 −x+1))/(x^2  +3))dx
$$\left.\mathrm{1}\right){find}\:\int\:\frac{\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$$$\left.\mathrm{2}\right){calculate}\:\int_{\mathrm{0}} ^{\mathrm{1}} \:\frac{\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx} \\ $$
Commented by mathmax by abdo last updated on 23/Feb/20
1) we have x^2 −x+1 =(x−(1/2))^2  +(3/4)  changement x−(1/2)=((√3)/2)sh(t)  give I=∫ ((√(x^2 −x+1))/(x^2  +3))dx =∫  ((((√3)/(2 )) ch(t))/(((1/2)+((√3)/2)sh(t))^2  +3))×((√3)/2)ch(t)dt  =(3/4) ∫   ((ch^2 (t))/(3+(1/4)(1+2(√3)sh(t)+3sh^2 (t))))dt  =3 ∫  ((ch^2 (t))/(13+2(√3)sh(t)+3sh^2 t))dt  =3 ∫  (((1+ch(2t))/2)/(13+2(√3)sh(t)+3×((ch(2t)−1)/2)))dt  =3∫  ((1+ch(2t))/(26+4(√3)sh(t)+3ch(2t)−3))dt  =3 ∫  ((1+((e^(2t)  +e^(−2t) )/2))/(23+4(√3)×((e^t −e^(−t) )/2)+3×((e^(2t) +e^(−2t) )/2)))dt  =3 ∫  ((2+e^(2t)  +e^(−2t) )/(46+4(√3)(e^t −e^(−t) )+3e^(2t) +3e^(−2t) ))dt  =_(e^t  =u)     3 ∫   ((2+u^2  +u^(−2) )/(46+4(√3)u −4(√3)u^(−1)  +3u^2  +3u^(−2) ))(du/u)  =3 ∫   ((2u^2  +u^4  +1)/(46u^3  +4(√3)u^4 −4(√3)u^2  +3u^5  +3u))du  =3 ∫  ((u^4  +2u^2  +1)/(u(3u^4  +4(√3)u^3  +46u^2 +3)))du  let decompose F(u) =((u^4  +2u^2  +1)/(u(3u^4  +4(√3)u^3  +46 u^2  +3)))  ...be continued...
$$\left.\mathrm{1}\right)\:{we}\:{have}\:{x}^{\mathrm{2}} −{x}+\mathrm{1}\:=\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\:\:{changement}\:{x}−\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right) \\ $$$${give}\:{I}=\int\:\frac{\sqrt{{x}^{\mathrm{2}} −{x}+\mathrm{1}}}{{x}^{\mathrm{2}} \:+\mathrm{3}}{dx}\:=\int\:\:\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}\:}\:{ch}\left({t}\right)}{\left(\frac{\mathrm{1}}{\mathrm{2}}+\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right)\right)^{\mathrm{2}} \:+\mathrm{3}}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ch}\left({t}\right){dt} \\ $$$$=\frac{\mathrm{3}}{\mathrm{4}}\:\int\:\:\:\frac{{ch}^{\mathrm{2}} \left({t}\right)}{\mathrm{3}+\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\mathrm{2}\sqrt{\mathrm{3}}{sh}\left({t}\right)+\mathrm{3}{sh}^{\mathrm{2}} \left({t}\right)\right)}{dt} \\ $$$$=\mathrm{3}\:\int\:\:\frac{{ch}^{\mathrm{2}} \left({t}\right)}{\mathrm{13}+\mathrm{2}\sqrt{\mathrm{3}}{sh}\left({t}\right)+\mathrm{3}{sh}^{\mathrm{2}} {t}}{dt}\:\:=\mathrm{3}\:\int\:\:\frac{\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{2}}}{\mathrm{13}+\mathrm{2}\sqrt{\mathrm{3}}{sh}\left({t}\right)+\mathrm{3}×\frac{{ch}\left(\mathrm{2}{t}\right)−\mathrm{1}}{\mathrm{2}}}{dt} \\ $$$$=\mathrm{3}\int\:\:\frac{\mathrm{1}+{ch}\left(\mathrm{2}{t}\right)}{\mathrm{26}+\mathrm{4}\sqrt{\mathrm{3}}{sh}\left({t}\right)+\mathrm{3}{ch}\left(\mathrm{2}{t}\right)−\mathrm{3}}{dt} \\ $$$$=\mathrm{3}\:\int\:\:\frac{\mathrm{1}+\frac{{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{2}}}{\mathrm{23}+\mathrm{4}\sqrt{\mathrm{3}}×\frac{{e}^{{t}} −{e}^{−{t}} }{\mathrm{2}}+\mathrm{3}×\frac{{e}^{\mathrm{2}{t}} +{e}^{−\mathrm{2}{t}} }{\mathrm{2}}}{dt} \\ $$$$=\mathrm{3}\:\int\:\:\frac{\mathrm{2}+{e}^{\mathrm{2}{t}} \:+{e}^{−\mathrm{2}{t}} }{\mathrm{46}+\mathrm{4}\sqrt{\mathrm{3}}\left({e}^{{t}} −{e}^{−{t}} \right)+\mathrm{3}{e}^{\mathrm{2}{t}} +\mathrm{3}{e}^{−\mathrm{2}{t}} }{dt} \\ $$$$=_{{e}^{{t}} \:={u}} \:\:\:\:\mathrm{3}\:\int\:\:\:\frac{\mathrm{2}+{u}^{\mathrm{2}} \:+{u}^{−\mathrm{2}} }{\mathrm{46}+\mathrm{4}\sqrt{\mathrm{3}}{u}\:−\mathrm{4}\sqrt{\mathrm{3}}{u}^{−\mathrm{1}} \:+\mathrm{3}{u}^{\mathrm{2}} \:+\mathrm{3}{u}^{−\mathrm{2}} }\frac{{du}}{{u}} \\ $$$$=\mathrm{3}\:\int\:\:\:\frac{\mathrm{2}{u}^{\mathrm{2}} \:+{u}^{\mathrm{4}} \:+\mathrm{1}}{\mathrm{46}{u}^{\mathrm{3}} \:+\mathrm{4}\sqrt{\mathrm{3}}{u}^{\mathrm{4}} −\mathrm{4}\sqrt{\mathrm{3}}{u}^{\mathrm{2}} \:+\mathrm{3}{u}^{\mathrm{5}} \:+\mathrm{3}{u}}{du} \\ $$$$=\mathrm{3}\:\int\:\:\frac{{u}^{\mathrm{4}} \:+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}\left(\mathrm{3}{u}^{\mathrm{4}} \:+\mathrm{4}\sqrt{\mathrm{3}}{u}^{\mathrm{3}} \:+\mathrm{46}{u}^{\mathrm{2}} +\mathrm{3}\right)}{du} \\ $$$${let}\:{decompose}\:{F}\left({u}\right)\:=\frac{{u}^{\mathrm{4}} \:+\mathrm{2}{u}^{\mathrm{2}} \:+\mathrm{1}}{{u}\left(\mathrm{3}{u}^{\mathrm{4}} \:+\mathrm{4}\sqrt{\mathrm{3}}{u}^{\mathrm{3}} \:+\mathrm{46}\:{u}^{\mathrm{2}} \:+\mathrm{3}\right)} \\ $$$$…{be}\:{continued}… \\ $$

Leave a Reply

Your email address will not be published. Required fields are marked *