Question Number 73181 by mathmax by abdo last updated on 07/Nov/19
$${calculate}\:\int_{\mathrm{1}} ^{\mathrm{3}\:} \:\frac{{x}−\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$
Commented by mathmax by abdo last updated on 07/Nov/19
![we have x^2 +x+1 =(x+(1/2))^2 +(3/4) changement x+(1/2)=((√3)/2)sh(t)give sh(t)=((2x+1)/( (√3))) ⇒ ∫_1 ^3 ((x−2)/( (√(x^2 +x+1))))dx =∫_(arsh((√3))) ^(argsh((7/( (√3))))) ((((√3)/2)sh(t)−(1/2)−2)/(((√3)/2)ch(t)))×((√3)/2)ch(t)dt =∫_(ln((√3)+(√4))) ^(ln((7/( (√3)))+(√(1+((49)/3) )))) (((√3)/2)sh(t)−(5/2))dt =((√3)/2) ∫_(ln((√3)+2)) ^(ln((7/( (√3) ))+(√(1+((49)/3))))) sh(t)dt−(5/2)(ln((7/( (√3)))+((√(52))/( (√3))))−ln(2+(√3))) =((√3)/4)[ e^t +e^(−t) ]_(ln(2+(√3))) ^(ln(((7+(√(52)))/( (√3))))) −(5/2){ln(((7+(√(52)))/( (√3))))−ln(2+(√3))} =((√3)/4){(((7+(√(52)))/( (√3))))+(((7+(√(52)))/( (√3))))^(−1) −(2+(√3))−(2+(√3))^(−1) } −(5/2){ln(((7+(√(52)))/( (√3))))−ln(2+(√3))}](https://www.tinkutara.com/question/Q73195.png)
$${we}\:{have}\:{x}^{\mathrm{2}} \:+{x}+\mathrm{1}\:=\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} \:+\frac{\mathrm{3}}{\mathrm{4}}\:\:{changement}\:{x}+\frac{\mathrm{1}}{\mathrm{2}}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right){give} \\ $$$${sh}\left({t}\right)=\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\:\Rightarrow \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \:\frac{{x}−\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}\:=\int_{{arsh}\left(\sqrt{\mathrm{3}}\right)} ^{{argsh}\left(\frac{\mathrm{7}}{\:\sqrt{\mathrm{3}}}\right)} \:\frac{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right)−\frac{\mathrm{1}}{\mathrm{2}}−\mathrm{2}}{\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ch}\left({t}\right)}×\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{ch}\left({t}\right){dt} \\ $$$$=\int_{{ln}\left(\sqrt{\mathrm{3}}+\sqrt{\mathrm{4}}\right)} ^{{ln}\left(\frac{\mathrm{7}}{\:\sqrt{\mathrm{3}}}+\sqrt{\mathrm{1}+\frac{\mathrm{49}}{\mathrm{3}}\:}\right)} \left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{sh}\left({t}\right)−\frac{\mathrm{5}}{\mathrm{2}}\right){dt} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\:\int_{{ln}\left(\sqrt{\mathrm{3}}+\mathrm{2}\right)} ^{{ln}\left(\frac{\mathrm{7}}{\:\sqrt{\mathrm{3}}\:}+\sqrt{\mathrm{1}+\frac{\mathrm{49}}{\mathrm{3}}}\right)} {sh}\left({t}\right){dt}−\frac{\mathrm{5}}{\mathrm{2}}\left({ln}\left(\frac{\mathrm{7}}{\:\sqrt{\mathrm{3}}}+\frac{\sqrt{\mathrm{52}}}{\:\sqrt{\mathrm{3}}}\right)−{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right) \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left[\:{e}^{{t}} \:+{e}^{−{t}} \right]_{{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)} ^{{ln}\left(\frac{\mathrm{7}+\sqrt{\mathrm{52}}}{\:\sqrt{\mathrm{3}}}\right)} −\frac{\mathrm{5}}{\mathrm{2}}\left\{{ln}\left(\frac{\mathrm{7}+\sqrt{\mathrm{52}}}{\:\sqrt{\mathrm{3}}}\right)−{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right\} \\ $$$$=\frac{\sqrt{\mathrm{3}}}{\mathrm{4}}\left\{\left(\frac{\mathrm{7}+\sqrt{\mathrm{52}}}{\:\sqrt{\mathrm{3}}}\right)+\left(\frac{\mathrm{7}+\sqrt{\mathrm{52}}}{\:\sqrt{\mathrm{3}}}\right)^{−\mathrm{1}} −\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)−\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)^{−\mathrm{1}} \right\} \\ $$$$−\frac{\mathrm{5}}{\mathrm{2}}\left\{{ln}\left(\frac{\mathrm{7}+\sqrt{\mathrm{52}}}{\:\sqrt{\mathrm{3}}}\right)−{ln}\left(\mathrm{2}+\sqrt{\mathrm{3}}\right)\right\} \\ $$
Answered by petrochengula last updated on 07/Nov/19
$${considerI}=\:\int\frac{{x}−\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}\left({x}−\mathrm{2}\right)}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}−\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{5}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx} \\ $$$$=\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{1}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\frac{\mathrm{1}}{\mathrm{4}}+\frac{\mathrm{3}}{\mathrm{4}}}}{dx} \\ $$$$=\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\frac{\mathrm{5}}{\mathrm{2}}\int\frac{\mathrm{1}}{\:\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}} \\ $$$${consider}\:\int\frac{\mathrm{1}}{\:\sqrt{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}}}{dx} \\ $$$$=\int\frac{\mathrm{1}}{\:\sqrt{\frac{\mathrm{3}}{\mathrm{4}}\left(\frac{\mathrm{4}}{\mathrm{3}}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\mathrm{1}\right)}}{dx} \\ $$$$=\frac{\mathrm{2}}{\:\sqrt{\mathrm{3}}}\int\frac{\mathrm{1}}{\:\sqrt{\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)^{\mathrm{2}} +\mathrm{1}}}{dx} \\ $$$${let}\:{sinh}\theta=\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\Rightarrow\sqrt{\mathrm{3}}{sinh}\theta=\mathrm{2}{x}+\mathrm{1}\Rightarrow\sqrt{\mathrm{3}}{cosh}\theta{d}\theta=\mathrm{2}\frac{{dx}}{{d}\theta}\Rightarrow{dx}=\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}{cosh}\theta{d}\theta={dx} \\ $$$$=\int{d}\theta \\ $$$$=\theta+{C} \\ $$$$={sinh}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{c} \\ $$$$\int\frac{{x}−\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}=\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}−\frac{\mathrm{5}}{\mathrm{2}}{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C} \\ $$$$\int_{\mathrm{1}} ^{\mathrm{3}} \frac{{x}−\mathrm{2}}{\:\sqrt{{x}^{\mathrm{2}} +{x}+\mathrm{1}}}{dx}=\sqrt{\mathrm{13}}−\frac{\mathrm{5}}{\mathrm{2}}{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{7}}{\:\sqrt{\mathrm{3}}}\right)−\sqrt{\mathrm{3}}+\frac{\mathrm{5}}{\mathrm{2}}{sinh}^{−\mathrm{1}} \left(\frac{\mathrm{3}}{\:\sqrt{\mathrm{3}}}\right) \\ $$
Commented by petrochengula last updated on 07/Nov/19
$${please}\:{check}\:{if}\:{it}\:{is}\:{correct} \\ $$