Question

How can we tell if:
\[\int_{-100}^{-10}\frac{x^{2}(x-1)^{2}+(x-1)^{2}+x^{2}}{(x^{3}-3x+1)^{2}}dx\] is rational? How can we tell if:
\[\int_{-100}^{-10}\frac{x^{2}(x-1)^{2}+(x-1)^{2}+x^{2}}{(x^{3}-3x+1)^{2}}dx\] is rational? @Mathematics

Answer 1

Been thinking about this for a bit <.< wolfram lets me know it is, even gives the answer, but no steps are available, and I dont see how a person without technological assistance can come to this conclusion.

Answer 2

maybe if you express it in partial fractions (which are rational)?

Answer 3

thats not a bad idea, i'll try it out :)

Answer 4

i am not sure this has to do with any calculation at all

Answer 5

thats what im hoping for really. you guys know how much i hate calculus <.< so im thinking im missing something.

Answer 6

Everything else in the problem has been working out too nicely for this to be the end of the road =/

Answer 7

my mind is a little slow the moment, but this is a rational function . that is the integrand is

Answer 8

yep, but the integral of a rational function isn't necessarily rational. you can get logarithms, or arcTans.

Answer 9

that is for sure.

Answer 10

Here's some wolfram stuff on it.
http://www.wolframalpha.com/input/?i=%5Cfrac%7Bx%5E%7B2%7D%28x-1%29%5E%7B2%7D%2B%28x-1%29%5E%7B2%7D%2Bx%5E%7B2%7D%7D%7B%28x%5E%7B3%7D-3x%2B1%29%5E%7B2%7D%7D

Answer 11

but unlike logs and arctan etc here is a rational function where the degree of the numerator is the degree of the denominator.

Answer 12

which gives me a clue that the integral should be a rational function, but i could be totally mistaken

Answer 13

the degree of denominator = 6
the degree of numerator = 4

Answer 14

oops! never mind...

Answer 15

the thing that is troublesome is that x^3 - 3x + 1 is irreducible over the rationals. it cannot be factored into rational linear factors

Answer 16

bah, i'll sleep on it i guess. Ive been looking at this problem for way too long -.- let me give you guys the original problem, maybe some context will help out.

Answer 17

yikes i don't even understand the first line!

Answer 18

the substitution seemed way too convenient, i think thats the right direction to go in.

Answer 19

maybe i should reword it then. I mean to say using the formula:\[u=\frac{ 1}{1-x}\], -10 becomes 1/11, and 1/11 becomes 101/100, and 101/100 becomes -10. i cant think of a decent way to say that though >.<

Answer 20

same thing with -100 and the lower limits.

Answer 21

i hate calculus

Answer 22

me too >.> this problems dumb <.<

Answer 23

i keep staring at the thing that says
"this substitution send this bunch of junk with x's into the identical bunch of junk with u's " and i know i am lost

Answer 24

ive been told i show too much work during the proof, and i dont need to show so much, so i chose not to put all the details. i have them on a piece of paper though, one sec.

Answer 25

it turned out that substitution didnt change the integrand (except for the dx part anyways). Thats too weird <.<