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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 ?

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    $\begingroup$ I couldn't imagine there's a better way to solve it. $\endgroup$ Commented Jul 17, 2012 at 20:52
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    $\begingroup$ 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. $\endgroup$ Commented Jul 17, 2012 at 20:53
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    $\begingroup$ Very nice indeed! $\endgroup$
    – copper.hat
    Commented Jul 17, 2012 at 21:06
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    $\begingroup$ @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 $B(x) = \dfrac{1}{x} - \dfrac{x^9+1}{x} = x^8$. $\endgroup$ Commented Jul 17, 2012 at 21:07
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    $\begingroup$ @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? $\endgroup$ Commented Jul 17, 2012 at 21:13

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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  • $\begingroup$ I believe that the solution to the integral is missing a $\log {|x|}$... but nice method! $\endgroup$
    – Matt Groff
    Commented Jul 23, 2012 at 14:00
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    $\begingroup$ Nothing is missing, note that this is $x^{-9}$, not $x^9$: $$-1/9 \log |1+x^{-9}|=1/9\log \left|\frac{x^9}{x^9+1}\right|=\log |x|-1/9\log |x^9+1|$$ $\endgroup$ Commented Jul 23, 2012 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 $$\int\frac{\mathrm dx}{x^{10}+x}=\ln |x|-\frac{\ln \left|1+x^{9}\right|}{9}+C$$

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