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Question Number 63404 by mathmax by abdo last updated on 03/Jul/19

1) find  ∫    ((x+1)/(x^3 −3x −2))dx  2) calculate  ∫_4 ^(+∞)      ((x+1)/(x^3 −3x +2))dx

$$\left.\mathrm{1}\right)\:{find}\:\:\int\:\:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{3}{x}\:−\mathrm{2}}{dx} \\ $$$$\left.\mathrm{2}\right)\:{calculate}\:\:\int_{\mathrm{4}} ^{+\infty} \:\:\:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{3}{x}\:+\mathrm{2}}{dx} \\ $$

Commented by MJS last updated on 03/Jul/19

− in 1)  + in 2)  is this right?

$$\left.−\:\mathrm{in}\:\mathrm{1}\right) \\ $$$$\left.+\:\mathrm{in}\:\mathrm{2}\right) \\ $$$$\mathrm{is}\:\mathrm{this}\:\mathrm{right}? \\ $$

Commented by mathmax by abdo last updated on 03/Jul/19

2) the Q is calculate ∫_4 ^(+∞)    ((x+1)/(x^3 −3x−2))dx

$$\left.\mathrm{2}\right)\:{the}\:{Q}\:{is}\:{calculate}\:\int_{\mathrm{4}} ^{+\infty} \:\:\:\frac{{x}+\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{2}}{dx}\: \\ $$

Answered by MJS last updated on 03/Jul/19

∫((x+1)/(x^3 −3x−2))dx=∫(dx/((x−2)(x+1)))=  =(1/3)∫(dx/(x−2))−(1/3)∫(dx/(x+1))=(1/3)ln ∣((x−2)/(x+1))∣ +C    ∫((x+1)/(x^3 −3x+2))dx=∫((x+1)/((x−1)^2 (x+2)))dx=  =(2/3)∫(dx/((x−1)^2 ))+(1/9)∫(dx/(x−1))−(1/9)∫(dx/(x+2))=  =−(2/(3(x−1)))+(1/9)ln ∣((x−1)/(x+2))∣ +C

$$\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{3}{x}−\mathrm{2}}{dx}=\int\frac{{dx}}{\left({x}−\mathrm{2}\right)\left({x}+\mathrm{1}\right)}= \\ $$$$=\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}−\mathrm{2}}−\frac{\mathrm{1}}{\mathrm{3}}\int\frac{{dx}}{{x}+\mathrm{1}}=\frac{\mathrm{1}}{\mathrm{3}}\mathrm{ln}\:\mid\frac{{x}−\mathrm{2}}{{x}+\mathrm{1}}\mid\:+{C} \\ $$$$ \\ $$$$\int\frac{{x}+\mathrm{1}}{{x}^{\mathrm{3}} −\mathrm{3}{x}+\mathrm{2}}{dx}=\int\frac{{x}+\mathrm{1}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} \left({x}+\mathrm{2}\right)}{dx}= \\ $$$$=\frac{\mathrm{2}}{\mathrm{3}}\int\frac{{dx}}{\left({x}−\mathrm{1}\right)^{\mathrm{2}} }+\frac{\mathrm{1}}{\mathrm{9}}\int\frac{{dx}}{{x}−\mathrm{1}}−\frac{\mathrm{1}}{\mathrm{9}}\int\frac{{dx}}{{x}+\mathrm{2}}= \\ $$$$=−\frac{\mathrm{2}}{\mathrm{3}\left({x}−\mathrm{1}\right)}+\frac{\mathrm{1}}{\mathrm{9}}\mathrm{ln}\:\mid\frac{{x}−\mathrm{1}}{{x}+\mathrm{2}}\mid\:+{C} \\ $$

Commented by mathmax by abdo last updated on 03/Jul/19

thank you sir.

$${thank}\:{you}\:{sir}. \\ $$

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