Question Number 38104 by maxmathsup by imad last updated on 21/Jun/18
$${find}\:\:\int_{\mathrm{1}} ^{+\infty} \:\:\:\:\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)\sqrt{{x}+\mathrm{3}}} \\ $$
Answered by MJS last updated on 23/Jun/18
![∫(dx/((x^2 +2)(√(x+3))))= [t=(√(x+3)) → dx=2(√(x+3))dt] =2∫(dt/(t^4 −6t^2 +11))= [t^4 −6t^2 +11=0 ⇒ ⇒ t_1 =−((√2)/2)((3+(√(11)))+(−3+(√(11)))i) t_2 =−((√2)/2)((3+(√(11)))−(−3+(√(11)))i) t_3 =((√2)/2)((3+(√(11)))+(−3+(√(11)))i) t_4 =((√2)/2)((3+(√(11)))−(−3+(√(11)))i) (t−t_1 )(t−t_2 )=t^2 +(√(6+2(√(11))))t+(√(11)) (t−t_3 )(t−t_4 )=t^2 −(√(6+2(√(11))))t+(√(11))] =2∫(dt/((t^2 −(√(6+2(√(11))))t+(√(11)))(t^2 +(√(6+2(√(11))))t+(√(11)))))= =2∫(dt/((t^2 −at+b)(t^2 +at+b)))= =2(∫(P_1 /(t^2 −at+b))dt+∫(P_2 /(t^2 +at+b))dt)= [this might be hard to find: we need P_1 (t^2 +at+b)+P_2 (t^2 −at+b)=1 set P_1 =a−t ∧ P_2 =a+t ⇒ ⇒ (a−t)(t^2 +at+b)+(a+t)(t^2 −at+b)=2ab ⇒ ⇒ P_1 =((a−t)/(2ab)); P_2 =((a+t)/(2ab))] =(1/(ab))(−∫((t−a)/(t^2 −at+b))dt+∫((t+a)/(t^2 +at+b))dt)= [t−a=((2t−a)/2)−(a/2); t+a=((2t+a)/2)+(a/2)] =(1/(2ab))(−∫((2t−a)/(t^2 −at+b))dt+a∫(dt/(t^2 −at+b))+∫((2t+a)/(t^2 +at+b))dt+a∫(dt/(t^2 +at+b)))= =(1/(2ab))(∫((2t+a)/(t^2 +at+b))dt−∫((2t−a)/(t^2 −at+b))dt)+(1/(2b))(∫(dt/(t^2 −at+b))+∫(dt/(t^2 +at+b)))= =(1/(2ab))ln∣((t^2 +at+b)/(t^2 −at+b))∣+(1/(2b))(∫(dt/((t−(a/2))^2 +b−(a^2 /4)))+∫(dt/((t+(a/2))^2 +b−(a^2 /4))))= [t=u(√(b−(a^2 /4)))±(a/2) ⇒ u=((t±(a/2))/( (√(b−(a^2 /4))))) → dt=(√(b−(a^2 /4)))du] =(1/(2ab))ln∣((t^2 +at+b)/(t^2 −at+b))∣+((√(4b−a^2 ))/(4b))(∫(du_1 /(u_1 ^2 +1))+∫(du_2 /(u_2 ^2 +1)))= =(1/(2ab))ln∣((t^2 +at+b)/(t^2 −at+b))∣+((√(4b−a^2 ))/(4b))(tan^(−1) u_1 +tan^(−1) u_2 )= =(1/(2ab))ln∣((t^2 +at+b)/(t^2 −at+b))∣+((√(4b−a^2 ))/(4b))(tan^(−1) (((√(4b−a^2 ))/2)(t−(a/2)))+tan^(−1) (((√(4b−a^2 ))/2)(t+(a/2))))= =((√(11))/(44))(√(−3+(√(11))))ln∣((x+(√(2(3+(√(11)))(x+3)))+3+(√(11)))/(x−(√(2(3+(√(11)))(x+3)))+3+(√(11))))∣+ +((√(22))/(44))(√(−3+(√(11))))(tan^(−1) (((√2)/2)((√((−3+(√(11)))(x+3)))−1)+tan^(−1) (((√2)/2)((√((−3+(√(11)))(x+3)))+1))))+C](https://www.tinkutara.com/question/Q38212.png)
$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +\mathrm{2}\right)\sqrt{{x}+\mathrm{3}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\left[{t}=\sqrt{{x}+\mathrm{3}}\:\rightarrow\:{dx}=\mathrm{2}\sqrt{{x}+\mathrm{3}}{dt}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{11}}= \\ $$$$\:\:\:\:\:\left[{t}^{\mathrm{4}} −\mathrm{6}{t}^{\mathrm{2}} +\mathrm{11}=\mathrm{0}\:\Rightarrow\right. \\ $$$$\:\:\:\:\:\:\Rightarrow\:{t}_{\mathrm{1}} =−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\left(\mathrm{3}+\sqrt{\mathrm{11}}\right)+\left(−\mathrm{3}+\sqrt{\mathrm{11}}\right)\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{t}_{\mathrm{2}} =−\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\left(\mathrm{3}+\sqrt{\mathrm{11}}\right)−\left(−\mathrm{3}+\sqrt{\mathrm{11}}\right)\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{t}_{\mathrm{3}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\left(\mathrm{3}+\sqrt{\mathrm{11}}\right)+\left(−\mathrm{3}+\sqrt{\mathrm{11}}\right)\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:{t}_{\mathrm{4}} =\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\left(\mathrm{3}+\sqrt{\mathrm{11}}\right)−\left(−\mathrm{3}+\sqrt{\mathrm{11}}\right)\mathrm{i}\right) \\ $$$$\:\:\:\:\:\:\left({t}−{t}_{\mathrm{1}} \right)\left({t}−{t}_{\mathrm{2}} \right)={t}^{\mathrm{2}} +\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{11}}}{t}+\sqrt{\mathrm{11}} \\ $$$$\left.\:\:\:\:\:\:\left({t}−{t}_{\mathrm{3}} \right)\left({t}−{t}_{\mathrm{4}} \right)={t}^{\mathrm{2}} −\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{11}}}{t}+\sqrt{\mathrm{11}}\right] \\ $$$$=\mathrm{2}\int\frac{{dt}}{\left({t}^{\mathrm{2}} −\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{11}}}{t}+\sqrt{\mathrm{11}}\right)\left({t}^{\mathrm{2}} +\sqrt{\mathrm{6}+\mathrm{2}\sqrt{\mathrm{11}}}{t}+\sqrt{\mathrm{11}}\right)}= \\ $$$$=\mathrm{2}\int\frac{{dt}}{\left({t}^{\mathrm{2}} −{at}+{b}\right)\left({t}^{\mathrm{2}} +{at}+{b}\right)}= \\ $$$$=\mathrm{2}\left(\int\frac{{P}_{\mathrm{1}} }{{t}^{\mathrm{2}} −{at}+{b}}{dt}+\int\frac{{P}_{\mathrm{2}} }{{t}^{\mathrm{2}} +{at}+{b}}{dt}\right)= \\ $$$$\:\:\:\:\:\left[\mathrm{this}\:\mathrm{might}\:\mathrm{be}\:\mathrm{hard}\:\mathrm{to}\:\mathrm{find}:\right. \\ $$$$\:\:\:\:\:\:\mathrm{we}\:\mathrm{need}\:{P}_{\mathrm{1}} \left({t}^{\mathrm{2}} +{at}+{b}\right)+{P}_{\mathrm{2}} \left({t}^{\mathrm{2}} −{at}+{b}\right)=\mathrm{1} \\ $$$$\:\:\:\:\:\:\mathrm{set}\:{P}_{\mathrm{1}} ={a}−{t}\:\wedge\:{P}_{\mathrm{2}} ={a}+{t}\:\Rightarrow \\ $$$$\:\:\:\:\:\:\Rightarrow\:\left({a}−{t}\right)\left({t}^{\mathrm{2}} +{at}+{b}\right)+\left({a}+{t}\right)\left({t}^{\mathrm{2}} −{at}+{b}\right)=\mathrm{2}{ab}\:\Rightarrow \\ $$$$\left.\:\:\:\:\:\:\Rightarrow\:{P}_{\mathrm{1}} =\frac{{a}−{t}}{\mathrm{2}{ab}};\:{P}_{\mathrm{2}} =\frac{{a}+{t}}{\mathrm{2}{ab}}\right] \\ $$$$=\frac{\mathrm{1}}{{ab}}\left(−\int\frac{{t}−{a}}{{t}^{\mathrm{2}} −{at}+{b}}{dt}+\int\frac{{t}+{a}}{{t}^{\mathrm{2}} +{at}+{b}}{dt}\right)= \\ $$$$\:\:\:\:\:\left[{t}−{a}=\frac{\mathrm{2}{t}−{a}}{\mathrm{2}}−\frac{{a}}{\mathrm{2}};\:{t}+{a}=\frac{\mathrm{2}{t}+{a}}{\mathrm{2}}+\frac{{a}}{\mathrm{2}}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ab}}\left(−\int\frac{\mathrm{2}{t}−{a}}{{t}^{\mathrm{2}} −{at}+{b}}{dt}+{a}\int\frac{{dt}}{{t}^{\mathrm{2}} −{at}+{b}}+\int\frac{\mathrm{2}{t}+{a}}{{t}^{\mathrm{2}} +{at}+{b}}{dt}+{a}\int\frac{{dt}}{{t}^{\mathrm{2}} +{at}+{b}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ab}}\left(\int\frac{\mathrm{2}{t}+{a}}{{t}^{\mathrm{2}} +{at}+{b}}{dt}−\int\frac{\mathrm{2}{t}−{a}}{{t}^{\mathrm{2}} −{at}+{b}}{dt}\right)+\frac{\mathrm{1}}{\mathrm{2}{b}}\left(\int\frac{{dt}}{{t}^{\mathrm{2}} −{at}+{b}}+\int\frac{{dt}}{{t}^{\mathrm{2}} +{at}+{b}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ab}}\mathrm{ln}\mid\frac{{t}^{\mathrm{2}} +{at}+{b}}{{t}^{\mathrm{2}} −{at}+{b}}\mid+\frac{\mathrm{1}}{\mathrm{2}{b}}\left(\int\frac{{dt}}{\left({t}−\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +{b}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}+\int\frac{{dt}}{\left({t}+\frac{{a}}{\mathrm{2}}\right)^{\mathrm{2}} +{b}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\right)= \\ $$$$\:\:\:\:\:\left[{t}={u}\sqrt{{b}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}\pm\frac{{a}}{\mathrm{2}}\:\Rightarrow\:{u}=\frac{{t}\pm\frac{{a}}{\mathrm{2}}}{\:\sqrt{{b}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}}\:\rightarrow\:{dt}=\sqrt{{b}−\frac{{a}^{\mathrm{2}} }{\mathrm{4}}}{du}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ab}}\mathrm{ln}\mid\frac{{t}^{\mathrm{2}} +{at}+{b}}{{t}^{\mathrm{2}} −{at}+{b}}\mid+\frac{\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}{\mathrm{4}{b}}\left(\int\frac{{du}_{\mathrm{1}} }{{u}_{\mathrm{1}} ^{\mathrm{2}} +\mathrm{1}}+\int\frac{{du}_{\mathrm{2}} }{{u}_{\mathrm{2}} ^{\mathrm{2}} +\mathrm{1}}\right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ab}}\mathrm{ln}\mid\frac{{t}^{\mathrm{2}} +{at}+{b}}{{t}^{\mathrm{2}} −{at}+{b}}\mid+\frac{\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}{\mathrm{4}{b}}\left(\mathrm{tan}^{−\mathrm{1}} \:{u}_{\mathrm{1}} \:+\mathrm{tan}^{−\mathrm{1}} \:{u}_{\mathrm{2}} \right)= \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}{ab}}\mathrm{ln}\mid\frac{{t}^{\mathrm{2}} +{at}+{b}}{{t}^{\mathrm{2}} −{at}+{b}}\mid+\frac{\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}{\mathrm{4}{b}}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}{\mathrm{2}}\left({t}−\frac{{a}}{\mathrm{2}}\right)\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{4}{b}−{a}^{\mathrm{2}} }}{\mathrm{2}}\left({t}+\frac{{a}}{\mathrm{2}}\right)\right)\right)= \\ $$$$=\frac{\sqrt{\mathrm{11}}}{\mathrm{44}}\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}}\mathrm{ln}\mid\frac{{x}+\sqrt{\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{11}}\right)\left({x}+\mathrm{3}\right)}+\mathrm{3}+\sqrt{\mathrm{11}}}{{x}−\sqrt{\mathrm{2}\left(\mathrm{3}+\sqrt{\mathrm{11}}\right)\left({x}+\mathrm{3}\right)}+\mathrm{3}+\sqrt{\mathrm{11}}}\mid+ \\ $$$$\:\:\:\:\:+\frac{\sqrt{\mathrm{22}}}{\mathrm{44}}\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}}\left(\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\sqrt{\left(−\mathrm{3}+\sqrt{\mathrm{11}}\right)\left({x}+\mathrm{3}\right)}−\mathrm{1}\right)+\mathrm{tan}^{−\mathrm{1}} \left(\frac{\sqrt{\mathrm{2}}}{\mathrm{2}}\left(\sqrt{\left(−\mathrm{3}+\sqrt{\mathrm{11}}\right)\left({x}+\mathrm{3}\right)}+\mathrm{1}\right)\right)\right)\right)+{C} \\ $$
Commented by abdo mathsup 649 cc last updated on 23/Jun/18
$${thank}\:{you}\:{sir}\:{Mjs}. \\ $$
Commented by MJS last updated on 23/Jun/18
$$\mathrm{you}'\mathrm{re}\:\mathrm{welcome}\:\mathrm{as}\:\mathrm{always}. \\ $$$$\mathrm{I}'\mathrm{m}\:\mathrm{integral}−\mathrm{addicted}… \\ $$$$\mathrm{I}\:\mathrm{didn}'\mathrm{t}\:\mathrm{solve}\:\mathrm{it}\:\mathrm{in}\:\left[\mathrm{1};\:\infty\left[\:\mathrm{because}\:\mathrm{it}'\mathrm{s}\:\mathrm{got}\:\mathrm{no}\right.\right. \\ $$$$“\mathrm{nice}''\:\mathrm{solution}\:\mathrm{due}\:\mathrm{to}\:\mathrm{the}\:\mathrm{logarithmic}\:\mathrm{part} \\ $$$$\mathrm{but}\:\mathrm{it}'\mathrm{s}\:\approx.\mathrm{261683} \\ $$
Commented by MJS last updated on 23/Jun/18
$$\mathrm{but}\:\mathrm{in}\:\left[−\mathrm{3};\:\infty\left[\:\mathrm{it}'\mathrm{s}\:\frac{\sqrt{\mathrm{22}}}{\mathrm{44}}\sqrt{−\mathrm{3}+\sqrt{\mathrm{11}}}\pi\right.\right. \\ $$
Commented by math khazana by abdo last updated on 23/Jun/18
$${thank}\:{you}\:{sir}… \\ $$