Question Number 133123 by abdomsup last updated on 19/Feb/21
$${find}\:\int\:\:\frac{{x}^{\mathrm{2}} {dx}}{{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}} \\ $$
Answered by Ñï= last updated on 19/Feb/21
$$\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}+\mathrm{2}}{{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}\mid+\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dx}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}\mid+\frac{\mathrm{2}}{\mathrm{3}}\int\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +{x}−\mathrm{1}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}\mid+\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{2}{x}+\mathrm{1}+\mathrm{3}}{{x}^{\mathrm{2}} +{x}−\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}\mid+\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}^{\mathrm{2}} +{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\mid\frac{{x}+\frac{\mathrm{1}}{\mathrm{2}}−\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}}{{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}}\mid+{C} \\ $$
Commented by mathmax by abdo last updated on 19/Feb/21
$$\mathrm{thNks}\:\mathrm{sir}\: \\ $$