Question-140652

Question Number 140652 by mohammad17 last updated on 10/May/21

Answered by TheSupreme last updated on 10/May/21

x^3 −x^2 +x−1=(x−1)(x^2 +1) ∫((3−x)/(x^3 −x^2 +x−1))dx=∫(A/(x−1))+((Bx+C)/(x^2 +1))dx Ax^2 +A+Bx^2 −Bx+Cx−C=3−x A+B=0 −B+C=−1 A−C=3 A=1 , B=−1, C=−2 ∫(1/(x−1))dx−∫((x+2)/(x^2 +1))dx=ln∣x−1∣−(1/2)ln(x^2 +1)−2tan^(−1) (x)+c ∫(1/(x(x+1)^2 ))dx=∫(A/x)+((B(x+1)+C)/((x+1)^2 ))dx Ax^2 +2Ax+A+Bx^2 +Bx+Cx=1 A=1 2A+B+C=0 A+B=0 A=1, B=−1, C=−1 ∫(1/x)dx−∫(1/((x+1)))dx−∫(1/((x+1)^2 ))dx=ln∣x∣−ln∣x+1∣+(1/(x+1))+c

$${x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{x}−\mathrm{1}=\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +\mathrm{1}\right) \\ $$$$\int\frac{\mathrm{3}−{x}}{{x}^{\mathrm{3}} −{x}^{\mathrm{2}} +{x}−\mathrm{1}}{dx}=\int\frac{{A}}{{x}−\mathrm{1}}+\frac{{Bx}+{C}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx} \\ $$$${Ax}^{\mathrm{2}} +{A}+{Bx}^{\mathrm{2}} −{Bx}+{Cx}−{C}=\mathrm{3}−{x} \\ $$$${A}+{B}=\mathrm{0} \\ $$$$−{B}+{C}=−\mathrm{1} \\ $$$${A}−{C}=\mathrm{3} \\ $$$${A}=\mathrm{1}\:,\:{B}=−\mathrm{1},\:{C}=−\mathrm{2} \\ $$$$\int\frac{\mathrm{1}}{{x}−\mathrm{1}}{dx}−\int\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +\mathrm{1}}{dx}={ln}\mid{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{2}}{ln}\left({x}^{\mathrm{2}} +\mathrm{1}\right)−\mathrm{2}{tan}^{−\mathrm{1}} \left({x}\right)+{c} \\ $$$$ \\ $$$$\int\frac{\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}=\int\frac{{A}}{{x}}+\frac{{B}\left({x}+\mathrm{1}\right)+{C}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx} \\ $$$${Ax}^{\mathrm{2}} +\mathrm{2}{Ax}+{A}+{Bx}^{\mathrm{2}} +{Bx}+{Cx}=\mathrm{1} \\ $$$${A}=\mathrm{1} \\ $$$$\mathrm{2}{A}+{B}+{C}=\mathrm{0} \\ $$$${A}+{B}=\mathrm{0} \\ $$$${A}=\mathrm{1},\:{B}=−\mathrm{1},\:{C}=−\mathrm{1} \\ $$$$\int\frac{\mathrm{1}}{{x}}{dx}−\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)}{dx}−\int\frac{\mathrm{1}}{\left({x}+\mathrm{1}\right)^{\mathrm{2}} }{dx}={ln}\mid{x}\mid−{ln}\mid{x}+\mathrm{1}\mid+\frac{\mathrm{1}}{{x}+\mathrm{1}}+{c} \\ $$$$ \\ $$

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