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Question-84884

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Question Number 84884 by bshahid010@gmail.com last updated on 17/Mar/20
Commented by mathmax by abdo last updated on 17/Mar/20
A =∫ (dx/((2x−1)(√(x^2 +x+3)))) changement 2x−1=t give x=((t+1)/2) A =∫ (dt/(2t(√((((t+1)^2 )/4)+((t+1)/2)+3)))) =∫ (dt/(2t(√((t^2 +2t+1+2t+2+12)/4)))) =∫ (dt/(t(√(t^2 +4t+15)))) =∫ (dt/(t(√((t+2)^2 +11)))) =_(t+2=(√(11))sh(u)) ∫ (((√(11))ch(u)dt)/(((√(11))sh(u)−2)(√(11))ch(u))) =∫ (du/( (√(11))×((e^u −e^(−u) )/2)−2)) =∫ ((2du)/( (√(11))e^u −(√(11))e^(−u) −4)) =_(e^u =α) ∫ ((2dα)/(α((√(11))α−(√(11))α^(−1) −4))) =2 ∫ (dα/( (√(11))α^2 −4α−(√(11)))) Δ^′ =4+11 =15 ⇒α_1 =((2+(√(15)))/( (√(11)))) and α_2 =((2−(√(15)))/( (√(11)))) ⇒ A =2 ∫ (dα/( (√(11))(α−α_1 )(α−α_2 ))) =(2/( (√(11))×((2(√(15)))/( (√(11)))))) ∫((1/(α−α_1 ))−(1/(α−α_2 )))dα =(1/( (√(15))))ln∣((α−α_1 )/(α−α_2 ))∣ +C =(1/( (√(15))))ln∣((e^u −α_1 )/(e^u −α_2 ))∣ +C we have u=argsh(((t+2)/( (√(11))))) =argsh(((2x−1+2)/( (√(11))))) =ln(((2x+1)/( (√(11))))+(√(1+(((2x+1)/( (√(11)))))^2 ))) ⇒ A =(1/( (√(15))))ln∣((((2x+1)/( (√(11))))+(√(1+(((2x+1)/( (√(11)))))^2 ))−((2+(√(15)))/( (√(11)))))/(((2x+1)/( (√(11))))+(√(1+(((2x+1)/( (√(11)))))^2 −((2−(√(15)))/( (√(11))))))))∣ +C
$${A}\:=\int\:\:\frac{{dx}}{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{3}}}\:{changement}\:\mathrm{2}{x}−\mathrm{1}={t}\:{give}\:{x}=\frac{{t}+\mathrm{1}}{\mathrm{2}} \\ $$$${A}\:=\int\:\:\frac{{dt}}{\mathrm{2}{t}\sqrt{\frac{\left({t}+\mathrm{1}\right)^{\mathrm{2}} }{\mathrm{4}}+\frac{{t}+\mathrm{1}}{\mathrm{2}}+\mathrm{3}}}\:=\int\:\:\frac{{dt}}{\mathrm{2}{t}\sqrt{\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}+\mathrm{2}{t}+\mathrm{2}+\mathrm{12}}{\mathrm{4}}}} \\ $$$$=\int\:\:\frac{{dt}}{{t}\sqrt{{t}^{\mathrm{2}} \:+\mathrm{4}{t}+\mathrm{15}}}\:=\int\:\:\frac{{dt}}{{t}\sqrt{\left({t}+\mathrm{2}\right)^{\mathrm{2}} +\mathrm{11}}} \\ $$$$=_{{t}+\mathrm{2}=\sqrt{\mathrm{11}}{sh}\left({u}\right)} \:\:\:\:\:\int\:\:\frac{\sqrt{\mathrm{11}}{ch}\left({u}\right){dt}}{\left(\sqrt{\mathrm{11}}{sh}\left({u}\right)−\mathrm{2}\right)\sqrt{\mathrm{11}}{ch}\left({u}\right)} \\ $$$$=\int\:\:\frac{{du}}{\:\sqrt{\mathrm{11}}×\frac{{e}^{{u}} −{e}^{−{u}} }{\mathrm{2}}−\mathrm{2}}\:=\int\:\frac{\mathrm{2}{du}}{\:\sqrt{\mathrm{11}}{e}^{{u}} −\sqrt{\mathrm{11}}{e}^{−{u}} −\mathrm{4}} \\ $$$$=_{{e}^{{u}} \:=\alpha} \:\:\:\:\:\int\:\:\:\frac{\mathrm{2}{d}\alpha}{\alpha\left(\sqrt{\mathrm{11}}\alpha−\sqrt{\mathrm{11}}\alpha^{−\mathrm{1}} −\mathrm{4}\right)}\:=\mathrm{2}\:\int\:\:\frac{{d}\alpha}{\:\sqrt{\mathrm{11}}\alpha^{\mathrm{2}} −\mathrm{4}\alpha−\sqrt{\mathrm{11}}} \\ $$$$\Delta^{'} \:=\mathrm{4}+\mathrm{11}\:=\mathrm{15}\:\Rightarrow\alpha_{\mathrm{1}} =\frac{\mathrm{2}+\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{11}}}\:{and}\:\alpha_{\mathrm{2}} =\frac{\mathrm{2}−\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{11}}}\:\Rightarrow \\ $$$${A}\:=\mathrm{2}\:\int\:\:\frac{{d}\alpha}{\:\sqrt{\mathrm{11}}\left(\alpha−\alpha_{\mathrm{1}} \right)\left(\alpha−\alpha_{\mathrm{2}} \right)} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{11}}×\frac{\mathrm{2}\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{11}}}}\:\int\left(\frac{\mathrm{1}}{\alpha−\alpha_{\mathrm{1}} }−\frac{\mathrm{1}}{\alpha−\alpha_{\mathrm{2}} }\right){d}\alpha\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{15}}}{ln}\mid\frac{\alpha−\alpha_{\mathrm{1}} }{\alpha−\alpha_{\mathrm{2}} }\mid\:+{C} \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{15}}}{ln}\mid\frac{{e}^{{u}} −\alpha_{\mathrm{1}} }{{e}^{{u}} −\alpha_{\mathrm{2}} }\mid\:+{C}\:\:{we}\:{have}\:{u}={argsh}\left(\frac{{t}+\mathrm{2}}{\:\sqrt{\mathrm{11}}}\right)\:={argsh}\left(\frac{\mathrm{2}{x}−\mathrm{1}+\mathrm{2}}{\:\sqrt{\mathrm{11}}}\right) \\ $$$$={ln}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{11}}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{11}}}\right)^{\mathrm{2}} }\right)\:\Rightarrow \\ $$$${A}\:=\frac{\mathrm{1}}{\:\sqrt{\mathrm{15}}}{ln}\mid\frac{\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{11}}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{11}}}\right)^{\mathrm{2}} }−\frac{\mathrm{2}+\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{11}}}}{\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{11}}}+\sqrt{\mathrm{1}+\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{11}}}\right)^{\mathrm{2}} −\frac{\mathrm{2}−\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{11}}}}}\mid\:+{C}\: \\ $$$$ \\ $$$$ \\ $$
Answered by MJS last updated on 17/Mar/20
∫(dx/((2x−1)(√(x^2 +x+3))))= [t=(√(x^2 +x+3)) → ((2(√(x^2 +x+3)))/(2x+1))dt] =∫(dt/(2t^2 −((11)/2)−(√(4t^2 −11))))= [t=((√(11))/2)cosh ln u =(((√(11))(u^2 +1))/(4u)) ⇔ u=((2t+(√(4t^2 −11)))/( (√(11)))) → dt=((√(11(4t^2 −11)))/(2(2t+(√(4t^2 −11)))))du] =(2/( (√(11))))∫(du/(u^2 −(4/( (√(11))))u−1))= =(1/( (√(15))))∫(du/(u−((2+(√(15)))/( (√(11))))))−(1/( (√(15))))∫(du/(u−((2−(√(15)))/( (√(11)))))) and now it′s easy
$$\int\frac{{dx}}{\left(\mathrm{2}{x}−\mathrm{1}\right)\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{3}}}= \\ $$$$\:\:\:\:\:\left[{t}=\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{3}}\:\rightarrow\:\frac{\mathrm{2}\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{3}}}{\mathrm{2}{x}+\mathrm{1}}{dt}\right] \\ $$$$=\int\frac{{dt}}{\mathrm{2}{t}^{\mathrm{2}} −\frac{\mathrm{11}}{\mathrm{2}}−\sqrt{\mathrm{4}{t}^{\mathrm{2}} −\mathrm{11}}}= \\ $$$$\:\:\:\:\:\left[{t}=\frac{\sqrt{\mathrm{11}}}{\mathrm{2}}\mathrm{cosh}\:\mathrm{ln}\:{u}\:=\frac{\sqrt{\mathrm{11}}\left({u}^{\mathrm{2}} +\mathrm{1}\right)}{\mathrm{4}{u}}\:\Leftrightarrow\:{u}=\frac{\mathrm{2}{t}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} −\mathrm{11}}}{\:\sqrt{\mathrm{11}}}\:\rightarrow\:{dt}=\frac{\sqrt{\mathrm{11}\left(\mathrm{4}{t}^{\mathrm{2}} −\mathrm{11}\right)}}{\mathrm{2}\left(\mathrm{2}{t}+\sqrt{\mathrm{4}{t}^{\mathrm{2}} −\mathrm{11}}\right)}{du}\right] \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{11}}}\int\frac{{du}}{{u}^{\mathrm{2}} −\frac{\mathrm{4}}{\:\sqrt{\mathrm{11}}}{u}−\mathrm{1}}= \\ $$$$=\frac{\mathrm{1}}{\:\sqrt{\mathrm{15}}}\int\frac{{du}}{{u}−\frac{\mathrm{2}+\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{11}}}}−\frac{\mathrm{1}}{\:\sqrt{\mathrm{15}}}\int\frac{{du}}{{u}−\frac{\mathrm{2}−\sqrt{\mathrm{15}}}{\:\sqrt{\mathrm{11}}}} \\ $$$$\mathrm{and}\:\mathrm{now}\:\mathrm{it}'\mathrm{s}\:\mathrm{easy} \\ $$
Answered by TANMAY PANACEA last updated on 17/Mar/20
2x−1=(1/t)→2dx=((−1)/t^2 )dt 2x=(1+(1/t))→x=(1/2)(1+(1/t)) dx=((−dt)/(2t^2 )) ∫((−dt)/(2t^2 ×(1/t)(√((1/4)(1+(1/t))^2 +(1/2)(1+(1/t))+3)))) ∫((−dt)/(2t(√((1/4)(1+(2/t)+(1/t^2 ))+(1/2)+(1/(2t))+3)))) ((−1)/2)∫(dt/(t(√((1/4)+(1/(2t))+(1/(4t^2 ))+(1/2)+(1/(2t))+3)))) ((−1)/2)∫(dt/(t(√((t^2 +2t+1+2t^2 +2t+12t^2 )/(4t^2 ))))) ((−1)/2)∫((2tdt)/(t(√(15t^2 +4t+1)))) −∫(dt/( (√(15(t^2 +((4t)/(15))+(1/(15))))))) ((−1)/( (√(15))))∫(dt/( (√((t+(2/(15)))^2 +(1/(15))−(4/(15^2 )))))) ((−1)/( (√(15))))∫(dt/( (√((t+(2/(15)))^2 +(((√(11))/(15)))^2 )))) ((−1)/( (√(15))))×ln{(t+(2/(15)))+(√((t+(2/(15)))^2 +(((√(11))/(15)))^2 )) }+C now put t=(1/(2x−1))
$$\mathrm{2}{x}−\mathrm{1}=\frac{\mathrm{1}}{{t}}\rightarrow\mathrm{2}{dx}=\frac{−\mathrm{1}}{{t}^{\mathrm{2}} }{dt} \\ $$$$\mathrm{2}{x}=\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)\rightarrow{x}=\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right) \\ $$$${dx}=\frac{−{dt}}{\mathrm{2}{t}^{\mathrm{2}} } \\ $$$$\int\frac{−{dt}}{\mathrm{2}{t}^{\mathrm{2}} ×\frac{\mathrm{1}}{{t}}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{2}}\left(\mathrm{1}+\frac{\mathrm{1}}{{t}}\right)+\mathrm{3}}} \\ $$$$\int\frac{−{dt}}{\mathrm{2}{t}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}\left(\mathrm{1}+\frac{\mathrm{2}}{{t}}+\frac{\mathrm{1}}{{t}^{\mathrm{2}} }\right)+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{t}}+\mathrm{3}}} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}\sqrt{\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{1}}{\mathrm{2}{t}}+\frac{\mathrm{1}}{\mathrm{4}{t}^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}{t}}+\mathrm{3}}} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{{dt}}{{t}\sqrt{\frac{{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{1}+\mathrm{2}{t}^{\mathrm{2}} +\mathrm{2}{t}+\mathrm{12}{t}^{\mathrm{2}} }{\mathrm{4}{t}^{\mathrm{2}} }}} \\ $$$$\frac{−\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{tdt}}{{t}\sqrt{\mathrm{15}{t}^{\mathrm{2}} +\mathrm{4}{t}+\mathrm{1}}} \\ $$$$−\int\frac{{dt}}{\:\sqrt{\mathrm{15}\left({t}^{\mathrm{2}} +\frac{\mathrm{4}{t}}{\mathrm{15}}+\frac{\mathrm{1}}{\mathrm{15}}\right)}} \\ $$$$\frac{−\mathrm{1}}{\:\sqrt{\mathrm{15}}}\int\frac{{dt}}{\:\sqrt{\left({t}+\frac{\mathrm{2}}{\mathrm{15}}\right)^{\mathrm{2}} +\frac{\mathrm{1}}{\mathrm{15}}−\frac{\mathrm{4}}{\mathrm{15}^{\mathrm{2}} }}} \\ $$$$\frac{−\mathrm{1}}{\:\sqrt{\mathrm{15}}}\int\frac{{dt}}{\:\sqrt{\left({t}+\frac{\mathrm{2}}{\mathrm{15}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{11}}}{\mathrm{15}}\right)^{\mathrm{2}} }} \\ $$$$\frac{−\mathrm{1}}{\:\sqrt{\mathrm{15}}}×{ln}\left\{\left({t}+\frac{\mathrm{2}}{\mathrm{15}}\right)+\sqrt{\left({t}+\frac{\mathrm{2}}{\mathrm{15}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{11}}}{\mathrm{15}}\right)^{\mathrm{2}} }\:\right\}+{C} \\ $$$${now}\:{put}\:{t}=\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{1}} \\ $$

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