Question Number 154186 by daus last updated on 15/Sep/21
Answered by puissant last updated on 15/Sep/21
![Ω=∫((x^3 −7x^2 +8x+3)/(x^2 −7x+12))dx =∫x−((4x−3)/(x^2 −7x+12))dx = (1/2)x^2 −∫((4x−3)/(x^2 −7x+12))dx =(1/2)x^2 −2∫((2x−(3/2)−((11)/2)+((11)/2))/(x^2 −7x+12))dx =(1/2)x^2 −2∫((2x−7)/(x^2 −7x+12))dx−11∫(dx/(x^2 −7x+12)) =(1/2)x^2 −2ln∣x^2 −7x+12∣−11∫(dx/((x−(7/2))−((49)/4)+12)) =(1/2)x^2 −2ln∣x^2 −7x+12∣−11∫(dx/((1/4)[2(x−(7/2))]^2 +1)) u=2(x−(7/2)) → dx=(du/2) Ω=(1/2)x^2 −2ln∣x^2 −7x+12∣−22∫(du/(1+u^2 )) =(1/2)x^2 −2ln∣x^2 −7x+12∣−22arctan(u)+C.. ∴∵ Ω = (1/2)x^2 −2ln∣x^2 −7x+12∣−22arctan(2x−7)+C..](https://www.tinkutara.com/question/Q154190.png)
$$\Omega=\int\frac{{x}^{\mathrm{3}} −\mathrm{7}{x}^{\mathrm{2}} +\mathrm{8}{x}+\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}}{dx} \\ $$$$=\int{x}−\frac{\mathrm{4}{x}−\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}}{dx}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\int\frac{\mathrm{4}{x}−\mathrm{3}}{{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{2}\int\frac{\mathrm{2}{x}−\frac{\mathrm{3}}{\mathrm{2}}−\frac{\mathrm{11}}{\mathrm{2}}+\frac{\mathrm{11}}{\mathrm{2}}}{{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{2}\int\frac{\mathrm{2}{x}−\mathrm{7}}{{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}}{dx}−\mathrm{11}\int\frac{{dx}}{{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{2}{ln}\mid{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}\mid−\mathrm{11}\int\frac{{dx}}{\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)−\frac{\mathrm{49}}{\mathrm{4}}+\mathrm{12}} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{2}{ln}\mid{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}\mid−\mathrm{11}\int\frac{{dx}}{\frac{\mathrm{1}}{\mathrm{4}}\left[\mathrm{2}\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)\right]^{\mathrm{2}} +\mathrm{1}} \\ $$$${u}=\mathrm{2}\left({x}−\frac{\mathrm{7}}{\mathrm{2}}\right)\:\rightarrow\:{dx}=\frac{{du}}{\mathrm{2}} \\ $$$$\Omega=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{2}{ln}\mid{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}\mid−\mathrm{22}\int\frac{{du}}{\mathrm{1}+{u}^{\mathrm{2}} } \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{2}{ln}\mid{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}\mid−\mathrm{22}{arctan}\left({u}\right)+{C}.. \\ $$$$ \\ $$$$\therefore\because\:\Omega\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x}^{\mathrm{2}} −\mathrm{2}{ln}\mid{x}^{\mathrm{2}} −\mathrm{7}{x}+\mathrm{12}\mid−\mathrm{22}{arctan}\left(\mathrm{2}{x}−\mathrm{7}\right)+{C}.. \\ $$