Question

The integral of 3/(16x^2 -9) using partial fractions: please help!

Answer 1

Hi----I have tried this several different ways, and I have so far gotten to identify the unknown constants A and B as 6 and -6, respectively

Answer 2

\[\int\limits(3/(16x^2 - 9)) dx = \int\limits(1/2)/(4x+3)dx - \int\limits(1/2)/(4x - 3)dx\]

Answer 3

i got coefficient 1/12 not 1/2...

Answer 4

I had 3 = A(4x-3) + B(4x +3); my factors were 3/4 and -3/4. Is that wrong?

Answer 5

\[=1/12 \int\limits_{ }^{}d(4x+3)/4(4x+3) -1/12\int\limits_{ }^{}d(4x-3)/4(4x-3)=\]

Answer 6

can you tell me what you did to get that answer with d(4x +/- 3)?

Answer 7

I checked your answer:
6/(4x-3) -6/(4x-3) = (24x+18-25x+18)/(4x+3)(4x-3) = 36/(4x-3)(4x+3)
it should 3 in numerator, so you have to divide by 12
right?

Answer 8

So you use the 6/-6 immediately in the eqn, not solve like 3 = A(6) +B(0)?

Answer 9

\[\int\limits_{ }^{}dx/(4x+3) = 1/4 \int\limits_{ }^{}d(4x+3)/(4x+3)\]
I did substitution to have the same variable in d() & denominator

Answer 10

I just took the answer that you got...
But normally you have to solve:
A/((4x+1) +b/(4x-3) = ...
I got now the same numbers for A & B (+/- 1/2)
playing with it now...

Answer 11

yeah, see what happens with +/- 1/2; I don't know what is wrong

Answer 12

wait: A/(4x +3), right?

Answer 13

nothing is wrong. i just checked...
let me re-write it from the start - will take a few

Answer 14

okay

Answer 15

so, from the A/(4x-3) + B/(4x-3) = 3/(....)
we got A=1/2 & b=-1/2
I think we agreed on this one

Answer 16

yep

Answer 17

And then when you re-write the integration, you pull the +/- 1/2 outside

Answer 18

now we got:
\[=1/2 \int\limits_{ }^{}dx/(4x-3) - 1/2\int\limits_{ }^{}dx/(4x+3)=\]
\[=1/2 *1/4\int\limits_{ }^{}d(4x-3)/(4x-3)-1/2 * 1/4 \int\limits_{}^{}d(4x+3)/(4x+3)=\]=

Answer 19

So I don't understand the last line: how do you know to use u-sub?

Answer 20

1/8 ln/4x-3/ -1/8 ln/4x+3/ +const=1/8 ln/(4x-3)/(4x+3)/ +const

Answer 21

yes, you can do u substitution like:
u=4x+/-3
du=1/4 dx
I just didn't write it this way...

Answer 22

so we need to make the demonimator u before evaluating it as a natural log?

Answer 23

yes, you can re-write as:
\[=1/2 *1/4 \int\limits_{ }^{}du/u -....\]
it's the same thing if you compare what I wrote before

Answer 24

okay, so now I have (1/8) ln((4x+3)/(4x-3))...

Answer 25

it has to be absolute value inside ln & + const
this is the answer (should be :) ?)

Answer 26

got the absolute value, but the answer is wrong

Answer 27

wrong all together or just a coefficient(s)?

Answer 28

I think something is wrong with the natural log

Answer 29

wait: I swtiched the +/- sign, and now it is right. How did I get that wrong?

Answer 30

I noticed that this was the way you had it---how did you get the negative sign on top?

Answer 31

actually we had it right - see above:
1/2∗1/4∫d(4x−3)/(4x−3)−1/2∗1/4∫d(4x+3)/(4x+3)=...

which means that ln should be : (4x-3)/(4x+3)...

Answer 32

I think the confusion came from when defined A & B.
A=1/2 gave me: 1/2(4x-3)
B=-1/2: 1/2(4x+3)

you may had it opposite...?

Answer 33

Probably. Thanks so much for the help!

Answer 34

Good Luck!
You may have to "clean" our solution a bit... :)