Question Number 150079 by puissant last updated on 09/Aug/21
$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} } \\ $$
Answered by Ar Brandon last updated on 09/Aug/21
$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{{px}+{q}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}+\int\frac{{rx}+{s}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{dx} \\ $$$$\frac{\mathrm{1}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{{p}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)−\left({px}+{q}\right)\left(\mathrm{2}{x}+\mathrm{1}\right)}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }+\frac{{rx}+{s}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$${r}=\mathrm{0},\:{s}+{r}−\mathrm{2}{p}+{p}=\mathrm{0}\Rightarrow{s}={p},\:{r}+{s}−\mathrm{2}{q}−{p}+{p}=\mathrm{0}\Rightarrow{p}=\mathrm{2}{q} \\ $$$${p}−{q}+{s}=\mathrm{1}\Rightarrow{q}=\frac{\mathrm{1}}{\mathrm{3}},\:{p}=\frac{\mathrm{2}}{\mathrm{3}}={s} \\ $$$$\int\frac{{dx}}{\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }=\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}+\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{1}} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}{x}+\mathrm{1}}{\mathrm{3}\left({x}^{\mathrm{2}} +{x}+\mathrm{1}\right)}+\frac{\mathrm{4}}{\mathrm{3}\sqrt{\mathrm{3}}}\mathrm{arctan}\left(\frac{\mathrm{2}{x}+\mathrm{1}}{\:\sqrt{\mathrm{3}}}\right)+{C} \\ $$
Commented by puissant last updated on 09/Aug/21
$${propre}\:{merci}\:{bro}.. \\ $$
Commented by puissant last updated on 09/Aug/21