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finding value of indefinite integration $\displaystyle \int\frac{y^7-y^5+y^3-y}{y^{10}+1}dy$

$\displaystyle \int\frac{y^5(y^2-1)+y(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2-1)}{y^{10}+1}dy = \int\frac{(y^5+y)(y^2+1-2)}{y^{10}+1}dy$

$\displaystyle =\int\frac{y^5+y}{y^8-y^6+y^4-y^2+1}-2\int\frac{y^5+y}{y^{10}+1}dy$

i wan,t be able to proceed after that, could some help me with this

DXT
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3 Answers3

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Hint: We can write our integral as, $$I = \int \frac{y^7-y^5+y^3-y}{y^{10}+1} dy = \int \frac{y(y^6-y^4 + y^2-1)}{y^{10}+1} dy = \int \frac{y(y^6-y^4+y^2-1)}{(y^2+1)(y^8-y^6+y^4-y^2+1)} dy$$ We can use partial fractions and get $$I = \frac{1}{5}\int \frac{4y^7-3y^5 +2y^3-y}{y^8-y^6+y^4-y^2+1} dy -\frac{4}{5} \int \frac{y}{y^2+1} dy =I_1-I_2$$ I hope you can take it from here.

  • I hope that you noticed that you answered came out 23 minutes ago and that, 9 minutes ago, was edited an answer which, initially, did not provide almost anything. In any manner, $+1$ for your answer. Cheers. – Claude Leibovici Dec 29 '16 at 14:01
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hint: put $u = y^2$, and note that $(y^5 + y)dy = y(y^4+1)dy = \dfrac{u^2+1}{2}du$. Would this enable you to continue?

DeepSea
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HINT: $$y^{10}+1= \left( {y}^{2}+1 \right) \left( {y}^{8}-{y}^{6}+{y}^{4}-{y}^{2}+1 \right) $$ further we get $$\frac{y^7-y^5+y^3-y}{y^{10}+1}=1/5\,{\frac {y \left( 4\,{y}^{6}-3\,{y}^{4}+2\,{y}^{2}-1 \right) }{{y} ^{8}-{y}^{6}+{y}^{4}-{y}^{2}+1}}-4/5\,{\frac {y}{{y}^{2}+1}} $$

Dr. Sonnhard Graubner
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