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3x-2-2x-4-7x-2-9x-2-dx-




Question Number 32785 by NECx last updated on 02/Apr/18
∫((3x^2 +2x−4)/(7x^2 −9x+2))dx
$$\int\frac{\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{4}}{\mathrm{7}{x}^{\mathrm{2}} −\mathrm{9}{x}+\mathrm{2}}{dx} \\ $$
Answered by Joel578 last updated on 02/Apr/18
((3x^2  + 2x − 4)/((7x − 2)(x −1))) = (A/(7x − 2)) + (B/(x − 1)) + C                                      = ((7Cx^2  + (A + 7B − 9C)x− (A + 2B − 2C))/((7x − 2)(x −1)))  C = (3/7),  B = (1/5), A = ((156)/(35))    I = ∫ ((3x^2  + 2x − 4)/(7x^2  − 9x + 2)) dx = ∫ ((156)/(35))((1/(7x − 2))) + (1/5)((1/(x −1))) + (3/7) dx     = ((156)/(245)) ln∣7x − 2∣ + (1/5) ln∣x − 1∣ + (3/7)x + C
$$\frac{\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{4}}{\left(\mathrm{7}{x}\:−\:\mathrm{2}\right)\left({x}\:−\mathrm{1}\right)}\:=\:\frac{{A}}{\mathrm{7}{x}\:−\:\mathrm{2}}\:+\:\frac{{B}}{{x}\:−\:\mathrm{1}}\:+\:{C} \\ $$$$\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:\:=\:\frac{\mathrm{7}{Cx}^{\mathrm{2}} \:+\:\left({A}\:+\:\mathrm{7}{B}\:−\:\mathrm{9}{C}\right){x}−\:\left({A}\:+\:\mathrm{2}{B}\:−\:\mathrm{2}{C}\right)}{\left(\mathrm{7}{x}\:−\:\mathrm{2}\right)\left({x}\:−\mathrm{1}\right)} \\ $$$${C}\:=\:\frac{\mathrm{3}}{\mathrm{7}},\:\:{B}\:=\:\frac{\mathrm{1}}{\mathrm{5}},\:{A}\:=\:\frac{\mathrm{156}}{\mathrm{35}} \\ $$$$ \\ $$$${I}\:=\:\int\:\frac{\mathrm{3}{x}^{\mathrm{2}} \:+\:\mathrm{2}{x}\:−\:\mathrm{4}}{\mathrm{7}{x}^{\mathrm{2}} \:−\:\mathrm{9}{x}\:+\:\mathrm{2}}\:{dx}\:=\:\int\:\frac{\mathrm{156}}{\mathrm{35}}\left(\frac{\mathrm{1}}{\mathrm{7}{x}\:−\:\mathrm{2}}\right)\:+\:\frac{\mathrm{1}}{\mathrm{5}}\left(\frac{\mathrm{1}}{{x}\:−\mathrm{1}}\right)\:+\:\frac{\mathrm{3}}{\mathrm{7}}\:{dx} \\ $$$$\:\:\:=\:\frac{\mathrm{156}}{\mathrm{245}}\:\mathrm{ln}\mid\mathrm{7}{x}\:−\:\mathrm{2}\mid\:+\:\frac{\mathrm{1}}{\mathrm{5}}\:\mathrm{ln}\mid{x}\:−\:\mathrm{1}\mid\:+\:\frac{\mathrm{3}}{\mathrm{7}}{x}\:+\:{C}\: \\ $$$$ \\ $$

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