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5x-4-4x-5-x-5-x-1-2-solve-the-intgration-




Question Number 19195 by vivek last updated on 06/Aug/17
∫((5x^4 +4x^5 )/((x^5 +x+1)^2 ))   solve the intgration
$$\int\frac{\mathrm{5}{x}^{\mathrm{4}} +\mathrm{4}{x}^{\mathrm{5}} }{\left({x}^{\mathrm{5}} +{x}+\mathrm{1}\right)^{\mathrm{2}} }\:\:\:{solve}\:{the}\:{intgration} \\ $$
Commented by vivek last updated on 07/Aug/17
please solve quetion
$${please}\:{solve}\:{quetion} \\ $$
Answered by ajfour last updated on 07/Aug/17
∫((5x^4 +4x^5 )/((x^5 +x+1)^2 ))dx  =∫((5x^4 +1)/((x^5 +x+1)^2 ))dx+∫((4x^5 −1)/((x^5 +x+1)^2 ))dx  =−(1/(x^5 +x+1))+C_1 +∫(((4x^3 −(1/x^2 )))/((x^4 +1−(1/x))^2 ))dx  = −(1/(x^5 +x+1)) −(1/((x^4 +1−(1/x)))) +C  I=−(1/((x^5 +x+1))) −(x/((x^5 +x−1))) +C .
$$\int\frac{\mathrm{5x}^{\mathrm{4}} +\mathrm{4x}^{\mathrm{5}} }{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\int\frac{\mathrm{5x}^{\mathrm{4}} +\mathrm{1}}{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx}+\int\frac{\mathrm{4x}^{\mathrm{5}} −\mathrm{1}}{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}+\mathrm{1}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} +\mathrm{x}+\mathrm{1}}+\mathrm{C}_{\mathrm{1}} +\int\frac{\left(\mathrm{4x}^{\mathrm{3}} −\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{2}} }\right)}{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)^{\mathrm{2}} }\mathrm{dx} \\ $$$$=\:−\frac{\mathrm{1}}{\mathrm{x}^{\mathrm{5}} +\mathrm{x}+\mathrm{1}}\:−\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{4}} +\mathrm{1}−\frac{\mathrm{1}}{\mathrm{x}}\right)}\:+\mathrm{C} \\ $$$$\mathrm{I}=−\frac{\mathrm{1}}{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}+\mathrm{1}\right)}\:−\frac{\mathrm{x}}{\left(\mathrm{x}^{\mathrm{5}} +\mathrm{x}−\mathrm{1}\right)}\:+\mathrm{C}\:. \\ $$

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