Question Number 32139 by Cheyboy last updated on 20/Mar/18
$${F}\boldsymbol{{ind}}\:\boldsymbol{{the}} \\ $$$$\int\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$
Commented by mrW2 last updated on 20/Mar/18
![∫ ((x+1)/(x^2 +x+1))dx =(1/2)∫ ((2x+1+1)/(x^2 +x+1))dx =(1/2)[∫ ((2x+1)/(x^2 +x+1))dx+∫(1/(x^2 +x+1))dx] =(1/2)[∫ (1/(x^2 +x+1))d(x^2 +x+1)+∫(1/(x^2 +x+1))dx] =(1/2)[ln (x^2 +x+1)+∫(1/((x+(1/2))^2 +(((√3)/2))^2 ))d(x+(1/2))] =(1/2)[ln (x^2 +x+1)+(2/(√3)) tan^(−1) ((2x+1)/(√3))]+C =(1/2)ln (x^2 +x+1)+(1/(√3)) tan^(−1) ((2x+1)/(√3))+C](Q32141.png)
$$\int\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\:\frac{\mathrm{2}{x}+\mathrm{1}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\int\:\frac{\mathrm{2}{x}+\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}+\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\int\:\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{d}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx}\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+\int\frac{\mathrm{1}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\left(\frac{\sqrt{\mathrm{3}}}{\mathrm{2}}\right)^{\mathrm{2}} }{d}\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)\right] \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\left[\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+\frac{\mathrm{2}}{\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}\right]+{C} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\mathrm{ln}\:\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)+\frac{\mathrm{1}}{\sqrt{\mathrm{3}}}\:\mathrm{tan}^{−\mathrm{1}} \frac{\mathrm{2}{x}+\mathrm{1}}{\sqrt{\mathrm{3}}}+{C} \\ $$
Commented by Cheyboy last updated on 20/Mar/18
$${Sir}\:{plzz}\:{can}\:{u}\:{explain}\:{this}\:{part} \\ $$$$\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{d}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right) \\ $$$$ \\ $$$$ \\ $$
Commented by Tinkutara last updated on 20/Mar/18
$$\frac{{d}}{{dx}}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\mathrm{2}{x}+\mathrm{1} \\ $$$${d}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)=\left(\mathrm{2}{x}+\mathrm{1}\right){dx} \\ $$$$\left(\mathrm{2}{x}+\mathrm{1}\right){dx}\:{is}\:{replaced}\:{by}\:{d}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right). \\ $$
Commented by Cheyboy last updated on 20/Mar/18
$${Ooh}!!\:{i}\:{have}\:{seen}\:{it}\:{Thank}\:{you}\:{for}\: \\ $$$${the}\:{clearification}. \\ $$$$ \\ $$