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How do I evaluate the following indefinite integral? $$ \int \frac{1}{x^{7} - x} ~ d{x}. $$ Could someone give me some advice as to what method I should use or the steps that I should take?


Note: The OP originally requested for help in evaluating $ \displaystyle \int \left( \frac{1}{x^{7}} - x \right) ~ d{x} $, which may not have been his/her actual intention.

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There is a trick. We have $$\frac{1}{x^7-x}=\frac{7x^6}{x^7-x} -\frac{7x^6-1}{x^7-x}.$$

The first function is $\dfrac{7x^5}{x^6-1}$. For integrating, there is an obvious substitution.

For the second function, there already is an obvious substitution.

One can invent many examples that yield to the same sort of trick.

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$\displaystyle \int\frac{1}{x^7-x}dx = \int\frac{1}{x^7.\left(1-\frac{1}{x^6}\right)}dx$

Put $\displaystyle \left(1-\frac{1}{x^6}\right) = t$ and $\displaystyle \frac{6}{x^7}dx = dt$

$\displaystyle = \frac{1}{6}\int\frac{1}{t}dt = \frac{1}{6}\ln \mid t \mid+C$

$\displaystyle = \frac{1}{6}\ln \left|\frac{x^6-1}{x^6}\right|+C$

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Factor the denom into $x{(x^3-1)}(x^{3}+1)$ and do a u-sub $x^{3}+1=u$ and it will go easy.

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  • $\begingroup$ Sorry, I was trying to edit, but it was done already, thanks. Anyway, this factorization with this u-sub really works well, it becomes a Partial Fraction Decomposition problem of three linear factors in the denom! $\endgroup$
    – imranfat
    Mar 25 '13 at 19:17

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