# Integral of $\frac{1}{( x^{2015} - x)}$

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.

• In theory you can integrate any rational function using partial fractions, but that might take a great lot of time. – Wojowu May 2 '16 at 8:26
• Hint: $\frac{dx}{x^{2015}-x} = \frac{x^{2013}dx}{(x^{2014}-1)x^{2014}}$ – achille hui May 2 '16 at 8:34

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