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Question Number 219506 by PragyanKhunte last updated on 27/Apr/25
Q. Integrate ((x^2 +x+1)/(2x^2 −x−3)).
$${Q}.\:{Integrate}\:\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{3}}. \\ $$$$ \\ $$
Answered by SdC355 last updated on 27/Apr/25
∫ ((x^2 +x+1)/(2x^2 −x−3))=∫ ((19)/(10(2x−3)))−(1/(5(x+1)))+(1/2) ((19)/(10)) ∫ (1/(2x−3))−(1/5)∫ (1/(x+1))+(1/2) ∫ 1 ((19)/(20)) ∫ (1/u) u=2x−3 =((19)/(20)) ln(2x−3) −(1/5)∫ (1/(x+1))=ln(x+1)=−(1/5)ln(x+1) (1/2) ∫ 1 = (1/2)x ∴((19)/(20)) ln(2x−3)−(1/5) ln(x+1)+(1/2)x+C
$$ \\ $$$$\int\:\:\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{3}}=\int\:\:\:\frac{\mathrm{19}}{\mathrm{10}\left(\mathrm{2}{x}−\mathrm{3}\right)}−\frac{\mathrm{1}}{\mathrm{5}\left({x}+\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{2}} \\ $$$$\frac{\mathrm{19}}{\mathrm{10}}\:\int\:\frac{\mathrm{1}}{\mathrm{2}{x}−\mathrm{3}}−\frac{\mathrm{1}}{\mathrm{5}}\int\:\:\frac{\mathrm{1}}{{x}+\mathrm{1}}+\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\mathrm{1} \\ $$$$\frac{\mathrm{19}}{\mathrm{20}}\:\int\:\:\frac{\mathrm{1}}{{u}}\:\:\:\:{u}=\mathrm{2}{x}−\mathrm{3}\:\:\:=\frac{\mathrm{19}}{\mathrm{20}}\:\mathrm{ln}\left(\mathrm{2}{x}−\mathrm{3}\right) \\ $$$$−\frac{\mathrm{1}}{\mathrm{5}}\int\:\:\frac{\mathrm{1}}{{x}+\mathrm{1}}=\mathrm{ln}\left({x}+\mathrm{1}\right)=−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\left({x}+\mathrm{1}\right) \\ $$$$\frac{\mathrm{1}}{\mathrm{2}}\:\int\:\mathrm{1}\:=\:\frac{\mathrm{1}}{\mathrm{2}}{x} \\ $$$$\therefore\frac{\mathrm{19}}{\mathrm{20}}\:\mathrm{ln}\left(\mathrm{2}{x}−\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{ln}\left({x}+\mathrm{1}\right)+\frac{\mathrm{1}}{\mathrm{2}}{x}+{C} \\ $$
Answered by mr W last updated on 27/Apr/25
∫((x^2 +x+1)/(2x^2 −x−3))dx =(1/2)∫((2x^2 −x−3+3x+5)/(2x^2 −x−3))dx =(1/2)∫(1+((3x+5)/(2x^2 −x−3)))dx =(x/2)+(1/2)∫((3x+5)/((2x−3)(x+1)))dx =(x/2)+(1/2)∫((A/(2x−3))+(B/(x+1)))dx (A+2B)x+(A−3B)=3x+5 ⇒A+2B=3, A−3B=5 ⇒B=−(2/5), A=((19)/5) =(x/2)+(1/2)∫[((19)/(5(2x−3)))−(2/(5(x+1)))]dx =(x/2)+((19)/(20))∫((d(2x−3))/((2x−3)))−(1/5)∫((d(x+1))/(x+1)) =(x/2)+((19)/(20))ln (2x−3)−(1/5)ln (x+1)+C
$$\int\frac{{x}^{\mathrm{2}} +{x}+\mathrm{1}}{\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{3}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{3}+\mathrm{3}{x}+\mathrm{5}}{\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{3}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\left(\mathrm{1}+\frac{\mathrm{3}{x}+\mathrm{5}}{\mathrm{2}{x}^{\mathrm{2}} −{x}−\mathrm{3}}\right){dx} \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{3}{x}+\mathrm{5}}{\left(\mathrm{2}{x}−\mathrm{3}\right)\left({x}+\mathrm{1}\right)}{dx} \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int\left(\frac{{A}}{\mathrm{2}{x}−\mathrm{3}}+\frac{{B}}{{x}+\mathrm{1}}\right){dx}\:\:\:\:\:\: \\ $$$$ \\ $$$$\left({A}+\mathrm{2}{B}\right){x}+\left({A}−\mathrm{3}{B}\right)=\mathrm{3}{x}+\mathrm{5} \\ $$$$\Rightarrow{A}+\mathrm{2}{B}=\mathrm{3},\:{A}−\mathrm{3}{B}=\mathrm{5}\:\Rightarrow{B}=−\frac{\mathrm{2}}{\mathrm{5}},\:{A}=\frac{\mathrm{19}}{\mathrm{5}} \\ $$$$ \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{1}}{\mathrm{2}}\int\left[\frac{\mathrm{19}}{\mathrm{5}\left(\mathrm{2}{x}−\mathrm{3}\right)}−\frac{\mathrm{2}}{\mathrm{5}\left({x}+\mathrm{1}\right)}\right]{dx}\:\:\:\:\:\: \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{19}}{\mathrm{20}}\int\frac{{d}\left(\mathrm{2}{x}−\mathrm{3}\right)}{\left(\mathrm{2}{x}−\mathrm{3}\right)}−\frac{\mathrm{1}}{\mathrm{5}}\int\frac{{d}\left({x}+\mathrm{1}\right)}{{x}+\mathrm{1}} \\ $$$$=\frac{{x}}{\mathrm{2}}+\frac{\mathrm{19}}{\mathrm{20}}\mathrm{ln}\:\left(\mathrm{2}{x}−\mathrm{3}\right)−\frac{\mathrm{1}}{\mathrm{5}}\mathrm{ln}\:\left({x}+\mathrm{1}\right)+{C} \\ $$

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