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Hello and thank you for taking interest in my question. This question came about when I was trying to integrate $$\int\frac{4x^5-1}{(x^5+x+1)^2} \,\mathrm{d}x$$ by the partial fraction method. I have found the solution by other methods and I have also seen the post in this site regarding this integral. I can't help but wonder if this can actually be integrated by using partial fractions.

So I tried to factor the denominator $x^5+x+1$ and found it to be $(x^3-x^2+1)(x^2+x+1)$. I am trying to rewrite $x^3-x^2+1$ as a product of a polynomial of degree 2 and a polynomial of degree $1$ but I cannot seem to calculate the root of $x^3-x^2+1$ i.e. I cannot find the solution of $x^3-x^2+1=0$. I graphed out $x^3-x^2+1$ and see it does have a root at $x=-0.755$. This may be stupid but may I please ask for help in algebraically finding the zero of this equation?

Thank You

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    $\begingroup$ That zero is $$\frac{1}{3} \left(1-\sqrt[3]{\frac{2}{25-3 \sqrt{69}}}-\sqrt[3]{\frac{1}{2} \left(25-3 \sqrt{69}\right)}\right).$$ I honestly don't think you want to use the method of partial fractions here :-) $\endgroup$ – Jyrki Lahtonen Mar 15 '18 at 6:15
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    $\begingroup$ Hint : Set $x=1/y$ $\endgroup$ – lab bhattacharjee Mar 15 '18 at 6:17
  • $\begingroup$ If this was given as an exercise, I would check whether a function of the form $f(x)=p(x)/(x^5+x+1)$ works. Differentiate that, and try to solve for $p(x)$. If that fails, I wouldn't touch this integral :-) $\endgroup$ – Jyrki Lahtonen Mar 15 '18 at 6:18
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Hint: Divide Numerator and Denominator by $x^2$.

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  • $\begingroup$ A nice trick! ${}$ $\endgroup$ – Jyrki Lahtonen Mar 15 '18 at 6:19
  • $\begingroup$ In this type of question, this is my first approach :) $\endgroup$ – kayush Mar 15 '18 at 6:20
  • $\begingroup$ Very elegant ! Can you tell me what you noticed in this question that made you do come to that solution ?? $\endgroup$ – lemniscate21 Mar 15 '18 at 7:12
  • $\begingroup$ @lemniscate21 In denominator there are 3 terms and in numerator there is 2 term, and $4 x^5 = 4 * x^3 * x^2$ thats why i divided by x^2 $\endgroup$ – kayush Mar 15 '18 at 7:14
  • $\begingroup$ I understand now. Thankyou ! :) $\endgroup$ – lemniscate21 Mar 15 '18 at 7:29

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