Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

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

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 ?

share|improve this question
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
show 5 more comments

1 Answer

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

share|improve this answer
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
add comment

Your Answer


By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.