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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 ?

Michael Hardy
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Frank
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    I couldn't imagine there's a better way to solve it. – David Mitra Jul 17 '12 at 20:52
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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. – Dylan Moreland Jul 17 '12 at 20:53
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    Very nice indeed! – copper.hat Jul 17 '12 at 21:06
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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 $B(x) = \dfrac{1}{x} - \dfrac{x^9+1}{x} = x^8$. – Robert Israel Jul 17 '12 at 21:07
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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? – Dylan Moreland Jul 17 '12 at 21:13
  • I think thats the best .you will get the same if u substitute $x=\frac {1}{y}$ . – Theorem Jul 17 '12 at 21:29
  • 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. – Michael Hardy Jul 17 '12 at 21:38
  • 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 $10$). – digital-Ink Jul 17 '12 at 21:45
  • Yowza! A clever prestidigitation! – ncmathsadist Jul 17 '12 at 22:00
  • What you did, or, equally fast and easy, the change of variable $t=1/x^9$. – Did Jul 19 '12 at 16:51

2 Answers2

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

Generic Human
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  • I believe that the solution to the integral is missing a $\log {|x|}$... but nice method! – Matt Groff Jul 23 '12 at 14:00
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    Nothing is missing, note that this is $x^{-9}$, not $x^9$: $$-1/9 \log |1+x^{-9}|=1/9\log \left|\frac{x^9}{x^9+1}\right|=\log |x|-1/9\log |x^9+1|$$ – Generic Human Jul 23 '12 at 14:11
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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 $$\int\frac{\mathrm dx}{x^{10}+x}=\ln |x|-\frac{\ln \left|1+x^{9}\right|}{9}+C$$

Venus
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