Yeah... I need help integrating it. It's 1/x^4 +1... I'm just... so lost.
+1 Karma to the person who helps >>
+1 Karma to the person who helps >>
What math is this again?
(level, that is)
(level, that is)
Calculus.
First you convert the 1/X^4 to X^-4, makes it so much easier.
Then integrate.
You get (-X^-3)/3 + X + C.
Or just 1/-3X^3 + X + C.
This is Calculus.
Edit: Whoops forgot a number.
Then integrate.
You get (-X^-3)/3 + X + C.
Or just 1/-3X^3 + X + C.
This is Calculus.
Edit: Whoops forgot a number.
Last edited by Smithereener (2007-06-11 19:34:10)
Yeah. I haven't got there yet.
>< Yeah, everyone's sprouting these short answers... -.-
I dunno though, cause some kid did it today, and it was like 2 pages... ><
I dunno though, cause some kid did it today, and it was like 2 pages... ><
Integration like that easy. If you know how to take a derivative using power rule, you should be able to integrate basic functions. Just do the opposite. In this case, you add one to the exponent and divide by the outcome. So X^-4 becomes X^-3 divided by the exponent, -3. The 1 turns into an X (X^0 ---> (X^0+1)/1) and you top it all off with a +C.joker8baller wrote:
>< Yeah, everyone's sprouting these short answers... -.-
I dunno though, cause some kid did it today, and it was like 2 pages... ><
-1/3x^3 + x + C
I give you that answer assuming that the denominator is only x^4, or is it x^4 + 1 in the denominator?
I give you that answer assuming that the denominator is only x^4, or is it x^4 + 1 in the denominator?
or +k depending on the way you were taught.
Yaocelotl your sig cracked me up
C is for constant. What does the K represent (the word, I know it's a constant )?Vilham wrote:
or +k depending on the way you were taught.
It's 1/(X^4 + 1 )Yaocelotl wrote:
-1/3x^3 + x + C
I give you that answer assuming that the denominator is only x^4, or is it x^4 + 1 in the denominator?
Well, if X^4+1 was the denominator, you'd have to probably use byparts. Unfortunately, I forgot how to do byparts (Summer mode already) and can't help you there. At least I'm pretty sure this is byparts.
Edit: Actually, wait. It might come out to be an arctan.
Edit: Actually, wait. It might come out to be an arctan.
Last edited by Smithereener (2007-06-11 19:50:36)
Man I feel stupid.
OK, now things change:joker8baller wrote:
It's 1/(X^4 + 1 )Yaocelotl wrote:
-1/3x^3 + x + C
I give you that answer assuming that the denominator is only x^4, or is it x^4 + 1 in the denominator?
that integral is the same as dx/x^4 + dx/x so the answer is -1/3x^3 + ln x + C
I don't think that's correct, you get 1/X^4 (Derivative of -1/3X^3) + 1/X (Derivative of ln x).Yaocelotl wrote:
OK, now things change:joker8baller wrote:
It's 1/(X^4 + 1 )Yaocelotl wrote:
-1/3x^3 + x + C
I give you that answer assuming that the denominator is only x^4, or is it x^4 + 1 in the denominator?
that integral is the same as dx/x^4 + dx/x so the answer is -1/3x^3 + ln x + C
Meant Derivatives .
Last edited by Smithereener (2007-06-11 20:08:55)
1+1=3?
dx/x = ln x + C, look it up .Smithereener wrote:
I don't think that's correct, you get 1/X^4 (Integral of -1/3X^3) + 1/X (Integral of ln x).Yaocelotl wrote:
OK, now things change:joker8baller wrote:
It's 1/(X^4 + 1 )
that integral is the same as dx/x^4 + dx/x so the answer is -1/3x^3 + ln x + C
Here's what came up on the Internet.
I have no clue on how that comes out, I haven't gotten too much into BC or Calc II material. Sorry.
@Yaocelotl
Haha, whoops I think I meant to say derivatives. When you derive the equation you came up with, you don't end up with the original function. I'm pretty sure that 1/X^4 + 1/X does not equal 1/(X^4 + 1). Or maybe I'm just fucked up in the head right now. I think I need a rest.
I have no clue on how that comes out, I haven't gotten too much into BC or Calc II material. Sorry.
@Yaocelotl
Haha, whoops I think I meant to say derivatives. When you derive the equation you came up with, you don't end up with the original function. I'm pretty sure that 1/X^4 + 1/X does not equal 1/(X^4 + 1). Or maybe I'm just fucked up in the head right now. I think I need a rest.
Oh yes, in fact, you are right, my bad.Smithereener wrote:
Here's what came up on the Internet.
http://img382.imageshack.us/img382/5611/calchq8.png
I have no clue on how that comes out, I haven't gotten too much into BC or Calc II material. Sorry.
@Yaocelotl
Haha, whoops I think I meant to say derivatives. When you derive the equation you came up with, you don't end up with the original function. I'm pretty sure that 1/X^4 + 1/X does not equal 1/(X^4 + 1). Or maybe I'm just fucked up in the head right now. I think I need a rest.
I think how you would solve it is trigonometric substitution, but i haven't tried it yet