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1-x-x-2-x-6-dx-




Question Number 172712 by ilhamQ last updated on 30/Jun/22
1. ∫(x/(x^2 −x−6)) dx=...
$$\mathrm{1}.\:\int\frac{{x}}{{x}^{\mathrm{2}} −{x}−\mathrm{6}}\:{dx}=… \\ $$
Answered by Ar Brandon last updated on 30/Jun/22
=(1/2)∫((2x−1)/(x^2 −x−6))dx+(1/2)∫(1/(x^2 −x−6))dx  =(1/2)∫((2x−1)/(x^2 −x−6))+(1/2)∫(1/((x−(1/2))^2 −((5/2))^2 ))dx  =ln(√(x^2 −x−6))+(1/(10))ln∣((2x−6)/(2x+4))∣+C
$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{dx}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{2}{x}−\mathrm{1}}{{x}^{\mathrm{2}} −{x}−\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{2}}\int\frac{\mathrm{1}}{\left({x}−\frac{\mathrm{1}}{\mathrm{2}}\right)^{\mathrm{2}} −\left(\frac{\mathrm{5}}{\mathrm{2}}\right)^{\mathrm{2}} }{dx} \\ $$$$=\mathrm{ln}\sqrt{{x}^{\mathrm{2}} −{x}−\mathrm{6}}+\frac{\mathrm{1}}{\mathrm{10}}\mathrm{ln}\mid\frac{\mathrm{2}{x}−\mathrm{6}}{\mathrm{2}{x}+\mathrm{4}}\mid+{C} \\ $$
Answered by thfchristopher last updated on 30/Jun/22
(x/(x^2 −x−6))  =(x/((x−3)(x+2)))=(A/(x−3))+(B/(x+2))  By solving, A=(3/5), B=(2/5)  ∴ ∫(x/(x^2 −x−6))dx  =(1/5)∫((3/(x−3))+(2/(x−2)))dx  =(3/5)∫(dx/(x−3))+(2/5)∫(dx/(x+2))  =(3/5)ln ∣x−3∣+(2/5)ln ∣x+2∣+C
$$\frac{{x}}{{x}^{\mathrm{2}} −{x}−\mathrm{6}} \\ $$$$=\frac{{x}}{\left({x}−\mathrm{3}\right)\left({x}+\mathrm{2}\right)}=\frac{{A}}{{x}−\mathrm{3}}+\frac{{B}}{{x}+\mathrm{2}} \\ $$$$\mathrm{By}\:\mathrm{solving},\:{A}=\frac{\mathrm{3}}{\mathrm{5}},\:{B}=\frac{\mathrm{2}}{\mathrm{5}} \\ $$$$\therefore\:\int\frac{{x}}{{x}^{\mathrm{2}} −{x}−\mathrm{6}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{5}}\int\left(\frac{\mathrm{3}}{{x}−\mathrm{3}}+\frac{\mathrm{2}}{{x}−\mathrm{2}}\right){dx} \\ $$$$=\frac{\mathrm{3}}{\mathrm{5}}\int\frac{{dx}}{{x}−\mathrm{3}}+\frac{\mathrm{2}}{\mathrm{5}}\int\frac{{dx}}{{x}+\mathrm{2}} \\ $$$$=\frac{\mathrm{3}}{\mathrm{5}}\mathrm{ln}\:\mid{x}−\mathrm{3}\mid+\frac{\mathrm{2}}{\mathrm{5}}\mathrm{ln}\:\mid{x}+\mathrm{2}\mid+{C} \\ $$

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