Question Number 189873 by mathocean1 last updated on 23/Mar/23
$$\int_{\mathrm{0}} ^{\:\mathrm{2}} \frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}=\:? \\ $$
Answered by Ar Brandon last updated on 23/Mar/23
![I=∫_0 ^2 ((x−1)/(x^2 +x+1))dx =(1/2)∫_0 ^2 ((2x+1)/(x^2 +x+1))dx−(3/2)∫_0 ^2 (dx/(x^2 +x+1)) =[(1/2)ln(x^2 +x+1)]_0 ^2 −(3/2)∫_0 ^2 (dx/((x+(1/2))^2 +(3/4))) =((ln7)/2)−(√3)[arctan(((2x+1)/( (√3))))]_0 ^2 =((ln7)/2)−(√3)(arctan((5/( (√3))))−arctan((1/( (√3))))) =((ln7)/2)+(π/(2(√3)))−(√3)arctan((5/( (√3))))](https://www.tinkutara.com/question/Q189875.png)
$${I}=\int_{\mathrm{0}} ^{\mathrm{2}} \frac{{x}−\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$$$\:\:\:=\frac{\mathrm{1}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}−\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$$\:\:\:=\left[\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} −\frac{\mathrm{3}}{\mathrm{2}}\int_{\mathrm{0}} ^{\mathrm{2}} \frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{3}}{\mathrm{4}}} \\ $$$$\:\:\:=\frac{\mathrm{ln7}}{\mathrm{2}}−\sqrt{\mathrm{3}}\left[\mathrm{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right]_{\mathrm{0}} ^{\mathrm{2}} \\ $$$$\:\:\:=\frac{\mathrm{ln7}}{\mathrm{2}}−\sqrt{\mathrm{3}}\left(\mathrm{arctan}\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}\right)−\mathrm{arctan}\left(\frac{\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)\right) \\ $$$$\:\:\:=\frac{\mathrm{ln7}}{\mathrm{2}}+\frac{\pi}{\mathrm{2}\sqrt{\mathrm{3}}}−\sqrt{\mathrm{3}}\mathrm{arctan}\left(\frac{\mathrm{5}}{\:\sqrt{\mathrm{3}}}\right) \\ $$