Question Number 35117 by math1967 last updated on 15/May/18
$$\int\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{4}} −\mathrm{3}{x}−\mathrm{2}}{dx} \\ $$
Answered by MJS last updated on 16/May/18
![∫((2x+3)/(x^4 −3x−2))dx=∫((2x+3)/((x^2 −x−1)(x^2 +x+2)))dx= =∫(dx/(x^2 −x−1))−∫(dx/(x^2 +x+2))= ∫(dx/(x^2 −x−1))=∫(4/(4x^2 −4x−4))dx= =∫(4/((2x−(√5)−1)(2x−(√5)+1)))dx= =((2(√5))/5)(∫(dx/((2x−(√5)−1)))−∫(dx/((2x−(√5)+1))))= =((√5)/5)ln(((∣2x−(√5)−1∣)/(∣2x−(√5)+1∣))) ∫(dx/(x^2 +x+2))=∫(dx/((x+(1/2))^2 +(7/4)))= [u=((√7)/7)(2x+1) → dx=((√7)/2)u] =((2(√7))/7)∫(du/(u^2 +1))=((2(√7))/7)arctan(u)= =((2(√7))/2)arctan(((√7)/7)(2x+1)) =((√5)/5)ln(((∣2x−(√5)−1∣)/(∣2x−(√5)+1∣)))−((2(√7))/2)arctan(((√7)/7)(2x+1))+C](https://www.tinkutara.com/question/Q35143.png)
$$\int\frac{\mathrm{2}{x}+\mathrm{3}}{{x}^{\mathrm{4}} −\mathrm{3}{x}−\mathrm{2}}{dx}=\int\frac{\mathrm{2}{x}+\mathrm{3}}{\left({x}^{\mathrm{2}} −{x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}+\mathrm{2}\right)}{dx}= \\ $$$$=\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}−\mathrm{1}}−\int\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{2}}= \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{{dx}}{{x}^{\mathrm{2}} −{x}−\mathrm{1}}=\int\frac{\mathrm{4}}{\mathrm{4}{x}^{\mathrm{2}} −\mathrm{4}{x}−\mathrm{4}}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\int\frac{\mathrm{4}}{\left(\mathrm{2}{x}−\sqrt{\mathrm{5}}−\mathrm{1}\right)\left(\mathrm{2}{x}−\sqrt{\mathrm{5}}+\mathrm{1}\right)}{dx}= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\sqrt{\mathrm{5}}}{\mathrm{5}}\left(\int\frac{{dx}}{\left(\mathrm{2}{x}−\sqrt{\mathrm{5}}−\mathrm{1}\right)}−\int\frac{{dx}}{\left(\mathrm{2}{x}−\sqrt{\mathrm{5}}+\mathrm{1}\right)}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\mathrm{ln}\left(\frac{\mid\mathrm{2}{x}−\sqrt{\mathrm{5}}−\mathrm{1}\mid}{\mid\mathrm{2}{x}−\sqrt{\mathrm{5}}+\mathrm{1}\mid}\right) \\ $$$$ \\ $$$$\:\:\:\:\:\:\:\:\:\:\int\frac{{dx}}{{x}^{\mathrm{2}} +{x}+\mathrm{2}}=\int\frac{{dx}}{\left({x}+\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} +\frac{\mathrm{7}}{\mathrm{4}}}= \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\left[{u}=\frac{\sqrt{\mathrm{7}}}{\mathrm{7}}\left(\mathrm{2}{x}+\mathrm{1}\right)\:\rightarrow\:{dx}=\frac{\sqrt{\mathrm{7}}}{\mathrm{2}}{u}\right] \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{7}}\int\frac{{du}}{{u}^{\mathrm{2}} +\mathrm{1}}=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{7}}\mathrm{arctan}\left({u}\right)= \\ $$$$\:\:\:\:\:\:\:\:\:\:=\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{2}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{7}}}{\mathrm{7}}\left(\mathrm{2}{x}+\mathrm{1}\right)\right) \\ $$$$ \\ $$$$=\frac{\sqrt{\mathrm{5}}}{\mathrm{5}}\mathrm{ln}\left(\frac{\mid\mathrm{2}{x}−\sqrt{\mathrm{5}}−\mathrm{1}\mid}{\mid\mathrm{2}{x}−\sqrt{\mathrm{5}}+\mathrm{1}\mid}\right)−\frac{\mathrm{2}\sqrt{\mathrm{7}}}{\mathrm{2}}\mathrm{arctan}\left(\frac{\sqrt{\mathrm{7}}}{\mathrm{7}}\left(\mathrm{2}{x}+\mathrm{1}\right)\right)+{C} \\ $$