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Question Number 165561 by LEKOUMA last updated on 03/Feb/22

I=∫(dx/(x^8 +x^6 ))  J=∫((1−x^7 )/(x(1+x^7 )))dx  Calculate I and J

$${I}=\int\frac{{dx}}{{x}^{\mathrm{8}} +{x}^{\mathrm{6}} } \\ $$$${J}=\int\frac{\mathrm{1}−{x}^{\mathrm{7}} }{{x}\left(\mathrm{1}+{x}^{\mathrm{7}} \right)}{dx} \\ $$$${Calculate}\:{I}\:{and}\:{J} \\ $$

Answered by MJS_new last updated on 05/Feb/22

∫(dx/(x^8 +x^6 ))=∫((1/x^6 )−(1/x^4 )+(1/x^2 )−(1/(x^2 +1)))dx=  =−(1/(5x^5 ))+(1/(3x^3 ))−(1/x)−arctan x =  =−((15x^4 −5x^2 +3)/(15x^5 ))−arctan x +C

$$\int\frac{{dx}}{{x}^{\mathrm{8}} +{x}^{\mathrm{6}} }=\int\left(\frac{\mathrm{1}}{{x}^{\mathrm{6}} }−\frac{\mathrm{1}}{{x}^{\mathrm{4}} }+\frac{\mathrm{1}}{{x}^{\mathrm{2}} }−\frac{\mathrm{1}}{{x}^{\mathrm{2}} +\mathrm{1}}\right){dx}= \\ $$$$=−\frac{\mathrm{1}}{\mathrm{5}{x}^{\mathrm{5}} }+\frac{\mathrm{1}}{\mathrm{3}{x}^{\mathrm{3}} }−\frac{\mathrm{1}}{{x}}−\mathrm{arctan}\:{x}\:= \\ $$$$=−\frac{\mathrm{15}{x}^{\mathrm{4}} −\mathrm{5}{x}^{\mathrm{2}} +\mathrm{3}}{\mathrm{15}{x}^{\mathrm{5}} }−\mathrm{arctan}\:{x}\:+{C} \\ $$

Answered by MJS_new last updated on 05/Feb/22

∫((1−x^7 )/(x(1+x^7 )))dx=−∫((x^7 −1)/(x(x+1)(x^6 −x^5 +x^4 −x^3 +x^2 −x+1)))dx=  =∫((1/x)−(2/(7(x+1)))−((2(6x^5 −5x^4 +4x^3 −3x^2 +2x−1))/(7(x^6 −x^5 +x^4 −x^3 +x^2 −x+1))))dx=  =ln x −(2/7)ln (x+1) −(2/7)ln (x^6 −x^5 +x^4 −x^3 +x^2 −x+1) =  =ln ∣x∣ −(2/7)ln ∣x^7 +1∣ +C

$$\int\frac{\mathrm{1}−{x}^{\mathrm{7}} }{{x}\left(\mathrm{1}+{x}^{\mathrm{7}} \right)}{dx}=−\int\frac{{x}^{\mathrm{7}} −\mathrm{1}}{{x}\left({x}+\mathrm{1}\right)\left({x}^{\mathrm{6}} −{x}^{\mathrm{5}} +{x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}{dx}= \\ $$$$=\int\left(\frac{\mathrm{1}}{{x}}−\frac{\mathrm{2}}{\mathrm{7}\left({x}+\mathrm{1}\right)}−\frac{\mathrm{2}\left(\mathrm{6}{x}^{\mathrm{5}} −\mathrm{5}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{3}} −\mathrm{3}{x}^{\mathrm{2}} +\mathrm{2}{x}−\mathrm{1}\right)}{\mathrm{7}\left({x}^{\mathrm{6}} −{x}^{\mathrm{5}} +{x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)}\right){dx}= \\ $$$$=\mathrm{ln}\:{x}\:−\frac{\mathrm{2}}{\mathrm{7}}\mathrm{ln}\:\left({x}+\mathrm{1}\right)\:−\frac{\mathrm{2}}{\mathrm{7}}\mathrm{ln}\:\left({x}^{\mathrm{6}} −{x}^{\mathrm{5}} +{x}^{\mathrm{4}} −{x}^{\mathrm{3}} +{x}^{\mathrm{2}} −{x}+\mathrm{1}\right)\:= \\ $$$$=\mathrm{ln}\:\mid{x}\mid\:−\frac{\mathrm{2}}{\mathrm{7}}\mathrm{ln}\:\mid{x}^{\mathrm{7}} +\mathrm{1}\mid\:+{C} \\ $$

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