Question

Partial Fraction Integral
1/(121-196x^2) dx

Answer 1

\[\frac{ 1 }{ 121-196x^2 } = \int\limits_{}^{} \frac{ A }{ 11+14x } + \frac{ B }{ 11-14x }\]

Answer 2

I then multiply everything by (11+14x)(11-14x).

Answer 3

\[1 = A(11-14x) + B(11+14x)\]

Answer 4

Then I rewrite it.

Answer 5

\[1 = 14(-A+B)x + 11(A+B)\] This is where I am stuck.

Answer 6

You don't really want to group the terms like that.
From this step:\[\large 1 = A(11-14x) + B(11+14x)\]

You have a couple of options,
You can substitute in \(\large x=\dfrac{11}{14}\)
which will allow you to solve for B.
After which you can find another value to sub in that will cancel out the B term, to solve for A.

If that method is too confusing, you can expand everything out and equate like terms.
This method will leave you with a system of equations.

Answer 7

From Mathematica:\[\frac{1}{121-196 x^2}=\frac{1}{22 (14 x+11)}-\frac{1}{22 (14 x-11)} \]

Answer 8

So the goal is to keep the letters factored out?

Answer 9

Umm, I guess we can try the second method if that explanation is a bit confusing.
Equating like terms from the step you last did,\[\large 1 = 14(-A+B)x + 11(A+B)\]
We can equate the constant values on each side,\[\large \color{royalblue}{1=11(A+B)}\]And we can equate the x^1 values on each side,\[\large 0x=14(-A+B)x\]Dividing through by x simplifies this second equation to,\[\large \color{royalblue}{0=14(-A+B)}\]

And from here, you have a system of 2 equations that can easily be solved.

Answer 10

Still struggling, now they we have equated the right side with the left. How do I solve for A and B?

Answer 11

Does 11 become A? and B become 14 based on position?

Answer 12

Based on position? I'm not quite sure what that means.
We can solve for A and B either through `substitution` or `elimination`.

So we have this system,\[1=11A+11B\]\[0=-14A+14B\]

Multiply the first equation by 14/11, giving us,\[\frac{14}{11}=14A+14B\]\[0=-14A+14B\]

From here, you can add the two equations together, and the A's will cancel out ~ Allowing you to solve for B.

Answer 13

This is the elimination method ^

Answer 14

Ahh OK, I get it now! I was trying to each one independently and not as a system. That explains so much! Thank you!

Answer 15

Ah I see :)