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I am trying to find the integral of $\dfrac{1}{( x^{2015} - x)}$.

Does anyone know how to do this? I can't possibly do a you substitutions right? Can't do partial fraction either.

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  • $\begingroup$ In theory you can integrate any rational function using partial fractions, but that might take a great lot of time. $\endgroup$
    – Wojowu
    Commented May 2, 2016 at 8:26
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    $\begingroup$ Hint: $\frac{dx}{x^{2015}-x} = \frac{x^{2013}dx}{(x^{2014}-1)x^{2014}}$ $\endgroup$ Commented May 2, 2016 at 8:34

2 Answers 2

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Hint :

Simplify it to $$\int \frac{1}{x^{2015}\left(1-\frac{1}{x^{2014}}\right)} \,dx$$ Then substitute $1-\frac{1}{x^{2014}}=t$

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    $\begingroup$ wow. how yall think like that. Thanks! solved it. $\endgroup$
    – RStyle
    Commented May 2, 2016 at 8:36
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Write the integral as $\int \frac{1}{x^{2015}(1-\frac{1}{x^{2014}})}\ dx$ . Then substitute : $ 1 - \frac{1}{x^{2014} }\ = u , du = \frac{2014}{x^{2015}}\ $ . Applying these will get you : $ \frac{1}{2014}\ \int \frac{1}{u}\ du = \frac{ln(u)}{2014}\ + c $ . Then substitute u back in and finally you get $ \frac{ln(1-x^{2014})}{2014}\ - ln(x) + c $ . Hope it was helpful !

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