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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 ?

  • 0
    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: $\begin{align} \int\frac{1}{x^{10}+x} dx=&\int\frac{x^{-10}}{1+x^{-9}} dx =-\frac 1 9 \log |1+x^{-9}| + C \end{align}$

  • 6
    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 $\int\frac{\mathrm dx}{x^{10}+x}=\ln |x|-\frac{\ln \left|1+x^{9}\right|}{9}+C$