Question

I think I'm throwing in the towel on this one. \[\int \frac{x^5-x}{x^8+1} dx\]
I know @amistre64 can do it though. :p This is @ganeshie8 's evil problem.

Answer 1

I have wrote this evil thing in several different ways...

Answer 2

\[\frac{x(x^4-1)}{x^8+1}=\frac{x(x^4-1)}{(x^4-\sqrt{2}x^2+1)(x^4+\sqrt{2}x^2+1)}\]
I was thinking this form might help because I peaked at the answer on wolfram. http://www.wolframalpha.com/input/?i=int%28%28x%5E5-x%29%2F%28x%5E8%2B1%29%2Cx%29

Answer 3

i did that when i was 3 ... ;)

Answer 4

partial fractions looks ugly

Answer 5

I was hoping for a trick.
I knew you did it at 3 so I knew you would know all the tricks to this one.

Answer 6

-x+x^5-x^8+- ....
---------
1+x^8 | -x+x^5
x-x^8
--------
x^5 - x^8
-x^5 -x^13
----------
-x^8-x^13

Answer 7

thats my approach, power series representation

Answer 8

im sure hartnn could whip up a laplace transform to put us all to shame :)

Answer 9

@hartnn

Answer 10

my power series approach migh tlook better with a split fraction ... same same tho

Answer 11

rational function.. so i think many ways to approach... myininaya's first method looks more straightforward to me.. :)

Answer 12

I don't like that way though. It is not cute at all.

Answer 13

haha cute :)

Answer 14

i hear fourier series are a god send as well .. cant say ive had much practice with them tho

Answer 15

I was looking to fall in love with this integral. I tried to get to know it and it does something disgusting.

Answer 16

math is evil :) a fickle thing she is

Answer 17

ganeshie8 said he loves this problem or likes
So I was just looking for the same relationship.

Answer 18

dividing x^4 top and bottom may bring back your good feelings i hope :

\[\int \frac{x^5-x}{x^8+1} dx = \int \dfrac{x-\frac{1}{x^3}}{x^4+\frac{1}{x^4}}dx\]

Answer 19

thats only half of the trick though.. the other half is in packing the denominator in an useful way..

Answer 20

noticing that derivative of 1/x^2 is -1/x^3

Answer 21

Yeah that is one of the ways I wrote it. In fact I divided by everything (at different times of course).
I will tried to play with this one in particular more though.

Answer 22

\[\large x^4 + \dfrac{1}{x^4} = \left( x^2+\dfrac{1}{x^2}\right)^2 - 2\]

Answer 23

well, moving the troublesome part into the numerator, usually helps.
i'd be tempted to try a u-sub with u=x^8+1

Answer 24

no more
i see
gosh
neat
i do love it again

Answer 25

well yes more
if people have other ways

Answer 26

But I see @ganeshie8 's way now

Answer 27

I really think I would have never thought of that

Answer 28

somebody in openstudy only showed me this solution... i think its eliassaab

Answer 29

but i feel there should be other more obvious and simple ways to work this...

Answer 30

I don't know about that.

Answer 31

However I could be wrong.

Answer 32

@SithsAndGiggles Always has tricks up his sleeves too.

Answer 33

wolfram has steps to solve integrals like this

Answer 34

but i think you have to subscribe

Answer 35

wolfram *shows* steps

Answer 36

And yeah I don't want to subscribe
I need my money

Answer 37

A substitution chain could make the original integral look a bit cleaner.
\[\begin{align*}\int\frac{x^5-x}{x^8+1}~dx&=\int\frac{x^5-x}{\left(x^4\right)^2+1}~dx\\\\
&=\int\frac{t^{5/4}-t^{1/4}}{t^2+1}\left(\frac{1}{4}t^{-3}{4}~dt\right)&\text{setting }t=x^4\\\\
&=\frac{1}{4}\int\frac{t^{2/4}-t^{-2/4}}{t^2+1}~dt\\\\
&=\frac{1}{4}\int\frac{t-1}{t^{1/2}\left(t^2+1\right)}~dt\\\\
&=\frac{1}{4}\int\frac{y^2-1}{y\left(y^4+1\right)}(2y~dy)&\text{setting }y=\sqrt t\\\\
&=\frac{1}{2}\int\frac{y^2-1}{y^4+1}~dy
\end{align*}\]

Answer 38

Second line should have \(\large\color{red}{t^{-3/4}}\)

Answer 39

\[\int\]

Answer 40

=]

Answer 41

Thank you all for your comments. I'm closing this but feel free to add anything you want. (I mean on this integral or other similar integrals)

Answer 42

its easy :P