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$$\int \frac{dx}{1+x^3}$$

I know it is to be solved by breaking the denominator as $(1+x)(1-x+x^2)$ and then do partialization of the fraction. But I wanted to know if there is any different method to solve it .

N. B. If the question is illogical, then I'll surely delete it. I am actually eager to know if there are alternative ways. Thanks.

Entrepreneur
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  • My two cents are that, in the end, the solution will be something in the form $A\ln(x+1)+B\ln(x^2-x+1)+C\arctan\frac{2x-1}{\sqrt3}$ with $A,B,C$ easily relatable to some $\alpha,\beta,\gamma$ such that $\frac{\alpha}{x+1}+\frac{\beta x+\gamma}{x^2-x+1}=\frac1{x^3+1}$. Possibly, more easily that it is to explain why they all turn out to be non-zero. –  Aug 27 '18 at 14:36
  • @SaucyO'Path $\alpha$ can easily be found . But $\beta$ and $\gamma$ are to be found by partialization. Any other way? – Entrepreneur Aug 27 '18 at 14:40
  • That's my point: $A,B,C$ are three fairly random numbers, so finding them in this instance is unreasonable unless one knows a general method that always finds them. But once you know it, it's the same as having partial fractions, because you can easily recover the partial fractions from the coefficients in the result. –  Aug 27 '18 at 14:45
  • For what my skepticism is worth, of course. –  Aug 27 '18 at 14:52

1 Answers1

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Try $$x=\frac{1-t}{1+t}$$ to get $$\int{\frac{1}{{{x}^{3}}+1}dx}=-\int{\frac{1+t}{1+3{{t}^{2}}}dt}$$

  • How did you think of it? Is there any technique? – Entrepreneur Aug 29 '18 at 19:24
  • this is called "the self similar substitution" which is effective in solving certain integrals, i give it a try and get the result!!! see this https://math.stackexchange.com/a/2870545/547564 –  Aug 29 '18 at 20:13