0
$\begingroup$

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.

$\endgroup$
  • $\begingroup$ In theory you can integrate any rational function using partial fractions, but that might take a great lot of time. $\endgroup$ – Wojowu May 2 '16 at 8:26
  • 3
    $\begingroup$ Hint: $\frac{dx}{x^{2015}-x} = \frac{x^{2013}dx}{(x^{2014}-1)x^{2014}}$ $\endgroup$ – achille hui May 2 '16 at 8:34
10
$\begingroup$

Hint :

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

$\endgroup$
  • 1
    $\begingroup$ wow. how yall think like that. Thanks! solved it. $\endgroup$ – RStyle May 2 '16 at 8:36
1
$\begingroup$

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 !

$\endgroup$

Your Answer

By clicking “Post Your Answer”, you agree to our terms of service, privacy policy and cookie policy

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