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Question Number 32785 by NECx last updated on 02/Apr/18

∫((3x^2 +2x−4)/(7x^2 −9x+2))dx

∫3x2+2x−47x2−9x+2dx

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

3x2+2x−4(7x−2)(x−1)=A7x−2+Bx−1+C=7Cx2+(A+7B−9C)x−(A+2B−2C)(7x−2)(x−1)C=37,B=15,A=15635I=∫3x2+2x−47x2−9x+2dx=∫15635(17x−2)+15(1x−1)+37dx=156245ln∣7x−2∣+15ln∣x−1∣+37x+C

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