Question

integrate x/(81^4-1)

Answer 1

The bottom is just a constant like that? 0_o

Answer 2

\[\large\rm \int\limits\frac{x}{c}dx\]Do you know how to integrate this?

Answer 3

yeah.. I was told to integrate using impartial fractions

Answer 4

i got \[\frac{ 1 }{ 81 }\ln \frac{ 9x^2-1 }{ 9x^2+1 }\]

Answer 5

Partial Fraction Decomposition?
Did you maybe post the problem incorrectly?
I don't see any x's in the bottom. Hmm

Answer 6

hmm

Answer 7

the 81^4 looks nice, probably part of something maybe, typo idk

Answer 8

you just have x/big number

Answer 9

\[\large\rm \int\limits \frac{x}{81x^4-1}dx\]This is what you meant to post maybe?

Answer 10

prolly

Answer 11

yes!

Answer 12

ooops hahaha

Answer 13

And you're instructed to Partial Fraction Decomp? :)
Ugh.

Answer 14

yeah i forgot about those for awhile till the DE class had to use them a bunch again

Answer 15

then figured out the calculator has command for decomp

Answer 16

SOS

Answer 17

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,\[\large\rm \frac{x}{(9x^2)^2-1}\quad=\frac{x}{(9x^2-1)(9x^2+1)}\]

Answer 18

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?

Answer 19

nope, I got this!

Answer 20

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

Answer 21

yes! i did this too

Answer 22

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)

Answer 23

use certain values for x that will make it easy, like put in x = -1/3

Answer 24

yes! I did that too

Answer 25

2 of those 3 will have a term go to zero,

x=-1/3

-1/3 = 0 + B*(-1 - 1)(1 + 1) + 0

Answer 26

you can do that method, or set up equations with undetermined coefficients on the like terms,

Answer 27

and you get systems of equations in the constant values, that you can solve

Answer 28

yes, but how come my answer was wrong?

Answer 29

I seriously can not figure this out for the life of me