Evaluating $\int \frac{1}{x^{7} - x} ~ d{x} $.

Question

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?

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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.

Answer 1

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.

Answer 2

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

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

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

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

Answer 3

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.

Answer 4

$$ \begin{aligned} \int \frac{1}{x^{7}-x} d x &=\int \frac{1}{x\left(x^{6}-1\right)} d x \\ &=\int\left(\frac{x^{5}}{x^{6}-1}-\frac{1}{x}\right) d x \\ &=\int \frac{x^{5} d x}{x^{6}-1}-\ln |x| \\ &=\frac{1}{6} \ln \left|x^{6}-1\right|-\ln |x|+C \\ &=\frac{1}{6} \ln \left|1-\frac{1}{x^{6}}\right|+C . \end{aligned} $$ Replacing 6 by $n$ and changing the sign yields the general integral $$\int \frac{1}{x\left(x^{n}\pm1\right)} d x=\mp \frac{1}{n} \ln \left|1\pm\frac{1}{x^{n}}\right|+C\tag*{}$$