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Question Number 133123 by abdomsup last updated on 19/Feb/21
find ∫  ((x^2 dx)/(x^3 −2x+1))
$${find}\:\int\:\:\frac{{x}^{\mathrm{2}} {dx}}{{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}} \\ $$
Answered by Ñï= last updated on 19/Feb/21
∫(x^2 /(x^3 −2x+1))dx  =(1/3)∫((3x^2 −2+2)/(x^3 −2x+1))dx  =(1/3)ln∣x^3 −2x+1∣+(2/3)∫(dx/((x−1)(x^2 +x−1)))  =(1/3)ln∣x^3 −2x+1∣+(2/3)∫((1/(x−1))−((x+2)/(x^2 +x−1)))dx  =(1/3)ln∣x^3 −2x+1∣+(2/3)ln∣x−1∣−(1/3)∫((2x+1+3)/(x^2 +x−1))dx  =(1/3)ln∣x^3 −2x+1∣+(2/3)ln∣x−1∣−(1/3)ln∣x^2 +x−1∣−(1/( (√5)))ln∣((x+(1/2)−(√(5/4)))/(x+(1/2)+(√(5/4))))∣+C
$$\int\frac{{x}^{\mathrm{2}} }{{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{3}{x}^{\mathrm{2}} −\mathrm{2}+\mathrm{2}}{{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}\mid+\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dx}}{\left({x}−\mathrm{1}\right)\left({x}^{\mathrm{2}} +{x}−\mathrm{1}\right)} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}\mid+\frac{\mathrm{2}}{\mathrm{3}}\int\left(\frac{\mathrm{1}}{{x}−\mathrm{1}}−\frac{{x}+\mathrm{2}}{{x}^{\mathrm{2}} +{x}−\mathrm{1}}\right){dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}\mid+\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{\mathrm{2}{x}+\mathrm{1}+\mathrm{3}}{{x}^{\mathrm{2}} +{x}−\mathrm{1}}{dx} \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}^{\mathrm{3}} −\mathrm{2}{x}+\mathrm{1}\mid+\frac{\mathrm{2}}{\mathrm{3}}{ln}\mid{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{\mathrm{3}}{ln}\mid{x}^{\mathrm{2}} +{x}−\mathrm{1}\mid−\frac{\mathrm{1}}{\:\sqrt{\mathrm{5}}}{ln}\mid\frac{{x}+\frac{\mathrm{1}}{\mathrm{2}}−\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}}{{x}+\frac{\mathrm{1}}{\mathrm{2}}+\sqrt{\frac{\mathrm{5}}{\mathrm{4}}}}\mid+{C} \\ $$
Commented by mathmax by abdo last updated on 19/Feb/21
thNks sir
$$\mathrm{thNks}\:\mathrm{sir}\: \\ $$

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