anonymous
  • anonymous
integrate x/(81^4-1)
Calculus1
  • Stacey Warren - Expert brainly.com
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SOLVED
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schrodinger
  • schrodinger
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zepdrix
  • zepdrix
The bottom is just a constant like that? 0_o
zepdrix
  • zepdrix
\[\large\rm \int\limits\frac{x}{c}dx\]Do you know how to integrate this?
anonymous
  • anonymous
yeah.. I was told to integrate using impartial fractions

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anonymous
  • anonymous
i got \[\frac{ 1 }{ 81 }\ln \frac{ 9x^2-1 }{ 9x^2+1 }\]
zepdrix
  • zepdrix
Partial Fraction Decomposition? Did you maybe post the problem incorrectly? I don't see any x's in the bottom. Hmm
DanJS
  • DanJS
hmm
DanJS
  • DanJS
the 81^4 looks nice, probably part of something maybe, typo idk
DanJS
  • DanJS
you just have x/big number
zepdrix
  • zepdrix
\[\large\rm \int\limits \frac{x}{81x^4-1}dx\]This is what you meant to post maybe?
DanJS
  • DanJS
prolly
anonymous
  • anonymous
yes!
anonymous
  • anonymous
ooops hahaha
zepdrix
  • zepdrix
And you're instructed to Partial Fraction Decomp? :) Ugh.
DanJS
  • DanJS
yeah i forgot about those for awhile till the DE class had to use them a bunch again
DanJS
  • DanJS
then figured out the calculator has command for decomp
anonymous
  • anonymous
SOS
zepdrix
  • zepdrix
So we can fiddle around with this 81 and x^4 a little bit, ya?\[\large\rm \frac{x}{(3x)^4-1}\]Which is more beneficial for us if we write it as a square I suppose.\[\large\rm \frac{x}{[(3x)^2]^2-1}\]Maybe that's a little confusing :) We'll write it the way you did, for now,\[\large\rm \frac{x}{(9x^2)^2-1}\]Apply your difference of squares identity,\[\large\rm \frac{x}{(9x^2)^2-1}\quad=\frac{x}{(9x^2-1)(9x^2+1)}\]
zepdrix
  • zepdrix
We can further apply this identity to the first set of brackets,\[\large\rm \frac{x}{([3x]^2-1)(9x^2+1)}\quad=\frac{x}{(3x-1)(3x+1)(9x^2+1)}\]And from here, we can start with the decomposition set up! :) Any confusion up to this point?
anonymous
  • anonymous
nope, I got this!
zepdrix
  • zepdrix
Here is what our "set up" will look like:\[\large\rm =\frac{A}{3x-1}+\frac{B}{3x+1}+\frac{Cx+D}{9x^2+1}\]What do you think? :d
anonymous
  • anonymous
yes! i did this too
DanJS
  • DanJS
equate numerators original to this, i believe, multiply all three those terms by (3x-1)(3x+1)(9x^2 + 1) x = A((3x+1)(9x^2+1) + B*(3x - 1)(9x^2+1) + (Cx+D)(3x-1)(3x+1)
DanJS
  • DanJS
use certain values for x that will make it easy, like put in x = -1/3
anonymous
  • anonymous
yes! I did that too
DanJS
  • DanJS
2 of those 3 will have a term go to zero, x=-1/3 -1/3 = 0 + B*(-1 - 1)(1 + 1) + 0
DanJS
  • DanJS
you can do that method, or set up equations with undetermined coefficients on the like terms,
DanJS
  • DanJS
and you get systems of equations in the constant values, that you can solve
anonymous
  • anonymous
yes, but how come my answer was wrong?
anonymous
  • anonymous
I seriously can not figure this out for the life of me

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