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$$\int \frac{1}{x^{10} + x}dx$$

My solution :

$$\begin{align*} \int\frac{1}{x^{10}+x}\,dx&=\int\left(\frac{x^9+1}{x^{10}+x}-\frac{x^9}{x^{10}+x}\right)\,dx\\ &=\int\left(\frac{1}{x}-\frac{x^8}{x^9+1}\right)\,dx\\ &=\ln|x|-\frac{1}{9}\ln|x^9+1|+C \end{align*}$$

Is there completely different way to solve it ?

  • 11
    I couldn't imagine there's a *better* way to solve it.2012-07-17
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    That's pretty clever! You could factorize the denominator completely over $\mathbf R$ and then use partial fractions, but that seems a lot less elegant.2012-07-17
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    Very nice indeed!2012-07-17
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    @DylanMoreland: Why factor completely when a partial factorization is enough? $x^{10} + x = x (x^9 + 1)$ and $x^9 + 1 = 1$ at $x=0$ so $\dfrac{1}{x^{10}+1} = \dfrac{1}{x} + \dfrac{B(x)}{x^9+1}$ where $B(x) = \dfrac{1}{x} - \dfrac{x^9+1}{x} = x^8$.2012-07-17
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    @Robert Well, that's what the OP does. Factoring completely is much worse, but it's what the procedure in a textbook would probably tell you to do, no?2012-07-17
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    I think thats the best .you will get the same if u substitute $x=\frac {1}{y}$ .2012-07-17
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    Probably there are _always_ completely different ways to do almost anything in mathematics. In this case: (1) Partial fractions as suggested above. (2) Probably contour integration in complex analysis. However, the way you've done it may be more efficient.2012-07-17
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    From my own experience, sometimes is best to forget about full partial fractions decomposition of rational function, at least when the polynomials involved have large degrees, and when the problem is a textbook exercise or some homework - you can always bet that there is a clever way to do it by a shorter, more convenient, decomposition, as is this case (which clearly works with any power, not only for $10$).2012-07-17
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    Yowza! A clever prestidigitation!2012-07-17
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    What you did, or, equally fast and easy, the change of variable $t=1/x^9$.2012-07-19

2 Answers 2

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Not really different, but even simpler: $$\begin{align} \int\frac{1}{x^{10}+x} dx=&\int\frac{x^{-10}}{1+x^{-9}} dx =-\frac 1 9 \log |1+x^{-9}| + C \end{align}$$

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    I believe that the solution to the integral is missing a $\log {|x|}$... but nice method!2012-07-23
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    Nothing is missing, note that this is $x^{-9}$, not $x^9$: $$-1/9 \log |1+x^{-9}|=1/9\log \left|\frac{x^9}{x^9+1}\right|=\log |x|-1/9\log |x^9+1|$$2012-07-23
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Let we generalise the problem with a slightly different way. Consider $$\int\frac{\mathrm dx}{x^n+x}$$ Making substitution $x=\frac{1}{y}$ and $z=y^{n-1}$ then reverse it back we get \begin{align} \int\frac{\mathrm dx}{x^n+x}&=-\int\frac{y^{n-2}}{1+y^{n-1}}\mathrm dy\\[9pt] &=-\frac{1}{n-1}\int\frac{\mathrm dz}{1+z}\\[9pt] &=-\frac{\ln |1+z|}{n-1}+C\\[9pt] &=\bbox[8pt,border:3px #FF69B4 solid]{\color{red}{\large\ln |x|-\frac{\ln \left|1+x^{n-1}\right|}{n-1}+C}} \end{align} In your case $$\int\frac{\mathrm dx}{x^{10}+x}=\ln |x|-\frac{\ln \left|1+x^{9}\right|}{9}+C$$