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

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

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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I couldn't imagine there's a better way to solve it. –  David Mitra Jul 17 '12 at 20:52
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
Very nice indeed! –  copper.hat Jul 17 '12 at 21:06
@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
@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

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}
I believe that the solution to the integral is missing a $\log {|x|}$... but nice method! –  Matt Groff Jul 23 '12 at 14:00
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