Question

solve :

Answer 1

\[\int\limits_{0}^{x}1/(1-x^{4})\]

Answer 2

you need to split this up into partial fractions first
i'll try this on pencil/paper - be back

Answer 3

ok, thanks

Answer 4

it comes to

1/2 int 1/(x^2 + 1)dx + 1/4 int (1-x) dx + 1/4 int 1/ (1+x) dx

Answer 5

thanks

Answer 6

sorry thats - 1/4 int 1 /(1-x)dx (the middle term)

Answer 7

the first integral is 1/2 arctan (x)
the others will be log to base e functions

Answer 8

some problem here because you are integrating from 0 to x and you have an x in the integrand. is this really the question?

Answer 9

oh yes -i hadn't noticed that

Answer 10

maybe the x outside the integral sign is a number matematik

Answer 11

maybe it is a c and not an x?

Answer 12

could be

Answer 13

is the question
\[\int_0^x\frac{dt}{1-t^4}\]?

Answer 14

it is x, because, this int was in series:
i had to find function for x^2/(1*2)-x^3/(2*3)+....

Answer 15

oh, sorry, not this series:

this one is correct
in x^(4n-3)/(4n-3)

Answer 16

\[\sum_{1}^{x}x ^{4n-3}/(4n-3)\]

Answer 17

maybe from 1 to infinity

Answer 18

yes:)

Answer 19

not sure what this is but this one
\[\sum_{k=1}^{\infty}\frac{x^n}{n}=-\ln(1-x)\] a well known one

Answer 20

i am trying to figure out what happens when you replace n by 4n-3

Answer 21

and i can't.
wolfram gives this
http://www.wolframalpha.com/input/?i=sum+n+%3D+1+to+infinity+x^%284n-3%29%2F%284n-3%29
but not sure how it gets it. what method are you using to find this?

Answer 22

i wrote x+x^5/5+x^9/9+....

so f ' (x)= 1+x^4+x^8+.... and this is 1/(1-x^4)
to find out what is f(x) I have to integrate this

Answer 23

and the answer should be1/4 ln((1+x)/(1-x))+1/2 arctg x

Answer 24

oooooooooooooh ok

Answer 25

then i am sorry i interrupted with my comment, jimmyrep had it. only thing is you just want the anti-derivative, not the integral from 0 to x
partial fractions will give it to you

Answer 26

\[\frac{1}{1-x^4}=\frac{1}{(1+x)(1-x)(1+x^2)}\] and so you need to break it up in to pieces

Answer 27

that is write
\[\frac{1}{1-x^4}=\frac{1}{4(1+x)}-\frac{1}{4(1-x)}+\frac{1}{2(1+x^2)}\] and integrate each piece separately. it is exactly what jimmyrep wrote

Answer 28

how did you come to this?

Answer 29

partial fraction decomposition. kind of a pain but no other way

Answer 30

\[\frac{1}{1-x^4}=\frac{a}{1-x}+\frac{b}{1+x}+\frac{cx+d}{1+x^2}\] solve for a, b, c, d

Answer 31

oh yes, thank you

Answer 32

yw. it will give i think exactly the answer you are looking for, since
\[\int\frac{dx}{1+x^2}=\tan^{-1}(x)\]
\[\int\frac{dx}{1-x}=\ln(1-x)\] etc