Partial Fractions?

Question

Question is indefinate integral { 10/[(x-1)(x^2+9)] dx }
So I did the work and got up to x^4(A+B) + x^3(-B+C) + x^2(2A+B-C+D) + x(-B+C-D+E)+A-C-E

Assuming I did the work correctly, I get
A-C-E = 10
-B+C-D+E = 0
2A+B-C+D = 0
-B+C = 0
A+B = 0

This isn't the answer to the question I'm really looking for, I'm just stuck, I can't figure out A,B,C,D,E. Anyone kind enough to help would be appreciated.

I should also mention I was able to work out some rules such as C = B, A = -B etc.. Nothing seems to work though.

Answer 1

When doing partial fractions, always look for a pattern: line up the variables carefully so you can see the pattern: in this case look at it this way:

A -C -E =10
-B+C -D +E =0
2A+B-C+D =0
-B+C =0
A +B =0

Adding all five equations gives you 4A=10 so A=5/2.

From (5), B=-5/2.

From (4), C=--5/2.

Similarly, (3)-(1)-(5)+(2)-(4) gives you 2E=-10 so E=-5. (Note (3)-(1)-(5) eliminates A.)

Next, (3)-(1)-(5) gives D+E=-10 so D=-5.

Unless the problem is really _evil_, try adding all the equations first to see if you get some answers for free like this one does. It is a good trick to keep in mind and can save you a ton of time in a testing situation. :)

Note: The equations looked lined up when I typed them, but apparently it lost it when it was submitted. Sorry about that...

Answer 2

You have 5 equations in 5 unknowns. If they're all independent, there's a unique solution. For convenience, label your equations 1 through 5. From #4 and #5, C = B and A = -B.

Substitute those into #1, #2, and #3 to get:

#6: -B - B - E = 10 ==> 2B + E = -10
#7: -B + B - D + E = 0 ==> D = E
#8: -2B + B - B + D = 0 ==> D = 2B

From #6 and #8, D + E = -10, but D and E are equal (#7), so

2D = -10 ==> D = E = -5

2B + E = 2B - 5 = -10 ==> B = -5/2

and finally C = B = -5/2, and A = -B = 5/2

Your answers for A through E are

5/2, -5/2, -5/2, -5, and -5

Answer 3

You're approaching the partial fraction wrong. Since the question is the integral of {10/[(x-1)(x^2+9)] dx}, you are looking for a partial fraction of the form A/(x-1) + (Bx+C)/(x^2+9). Then, you set the partial fraction equal to what's in the integral: 10/[(x-1)(x^2+9)] = A/(x-1) + (Bx+C)/(x^2+9). You can simplify this to 10 = A(x^2+9) + (Bx+C)(x-1). If you plug in three amounts for x, say x=1, x=2, x=3, you will find that A=1, B=C= -1. So, you want to find the integral of the partial fraction, that is, the integral of {1/(x-1) - x/(x^2+9) - 1/(x^2+9) dx}. These integrals are simple. The answer is ln|x-1| - 1/2*ln|x^2+9| - 1/3*arctan(x/3).

Answer 4

1)
Add the last two equations
-B+C = 0
A+B = 0

A + C = 0

2)
Add the second and third equations:

-B+C-D+E = 0
2A+B-C+D = 0
2A + E = 0
E = -2A

3)
Use E = -2A in the first equation:

A-C-E = 10 ----> A - C + 2A = 10

3A - C = 10



4)
Add the results from steps 1) and 3)
3A - C = 10
A + C = 0

4A = 10
A = (5/2)

5)
Find the other constants, using A = (5/2)

A+B = 0 -----> B = -(5/2)
-B+C = 0 -----> C = -(5/2)

2A+B-C+D = 0 -------> 2(5/2) - (5/2) + (5/2) + D = 0
D = -5

-B+C-D+E = 0 ----> (5/2) - (5/2) + 5 + E = 0
E = -5

Answer 5

(4) B=C
Substitute (4) in (3)
2A + D = 0 (6)
Substitute (4) in (2)
D=E (7)
(5) A=-B, so A=-C (because of (4))
Substitute (5) in (1)
2A -E = 10 (8)
Combine (6), (7), (8)
D=-5, E=-5, A=2.5
So B=-2.5 and C=-2.5

Answer 6

Been a while, but use your "rules a=-b" to substitute other equations until you have only one avriable. I didn't see anything easy here, maybe a program like mathematica could give you a numerical solution

Answer 7

If you would have asked this question about 35 years ago, I would have leaned over and asked my friend Bill. He'd know.