Standard Method Solution to Problem 11.7-3© 2001 by Karl Hahn |
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This page applies the standard method of partial fractions to find the solution. To see the Heaviside method applied to this same problem, click here.
The problem is to find the indefinite integral
11x3 - 107x + 108 |
Step 1: Set it up. The integrand is a rational function with a fourth degree polynomial in its denominator. So there will be four unknowns. You can call them, A, B, C, and D, although any four distinct symbols would have been ok. This denominator has repeated roots. Indeed the (x - 2) root is repeated three times. Recall how we set up repeated roots in the main text. Do this one that same way. Write that out, and then click here.
You should have
11x3 - 107x + 108 |
A B
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C D
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although any ordering of A, B, C, and D over the four binomials listed would be ok.
Step 2: Put all the partial fractions over the common denominator and
then arrange the terms according to powers of x. You will have to
multiply out each of the following:
(x + 7)(x - 2)2,
(x + 7)(x - 2), and
(x - 2)3.
Each of these will each be multiplied by one of the unknowns.
In addition, one of the unknowns will be multiplied by (x + 7)
alone. Figure out what goes with what and multiply it all out,
then click here.
To get the combined numerator on the right, you multiply A by (x - 2)3 = x3 - 6x2 + 12x - 8. You multiply B by (x + 7)(x - 2)2 = x3 + 3x2 - 24x + 28. You multiply C by (x + 7)(x - 2) = x2 + 5x - 14. You multiply D by (x + 7). Take the sum of those products, gather like powers of x, and you have
11x3 - 107x + 108 |
(A + B)x3
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(-6A + 3B + C)x2
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(12A - 24B + 5C + D)x
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(-8A + 28B - 14C + 7D)
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Step 3: Equate the numerators according to powers of x. This will give you four linear equations in A, B, C, and D. Go ahead and write them all out, then click here.
Your equations should be
11 = A + B
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equating coefficients of the x3 terms | ||
0 = -6A + 3B + C
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equating coefficients of the x2 terms | ||
-107 = 12A - 24B + 5C + D
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equating coefficients of the x terms | ||
108 = -8A + 28B - 14C + 7D
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equating coefficients of the constant terms |
Step 4: Solve the linear equation. This is not as bad as you think. Each step is simple. It's just that there are a lot of steps and it is easy to make arithmetic and sign errors. Try doing the Gaussian elimination process yourself. Then click here to see if you got the same results I did. Using a calculator helps a lot, but you still have to keep careful track of all your signs.
Here is the procedure for solving the linear equations in the example for A, B, C, and D by the method of Gaussian elimination. Each procedure tells you how to get from the current set of equations to the next set of equations. (Or you can go to QuickMath, enter in all four equations and the list of variables, and have it do it all for you, but what fun is that?)
| Equations | Procedure | |
A + B = 11
-6A + 3B + C = 0
12A - 24B + 5C + D = -107
-8A + 28B - 14C + 7D = 108
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Add 6 times first equation to second equation. Subtract 12 times first equation from third equation. Add 8 times first equation to fourth equation. | |
A + B = 11
9B + C = 66
-36B + 5C + D = -239
36B - 14C + 7D = 196
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Add 4 times second equation to third equation. Subtract 4 times second equation from fourth equation. | |
A + B = 11
9B + C = 66
9C + D = 25
-18C + 7D = -68
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Add 2 times third equation to fourth equation. Then divide fourth equation by 9 | |
A + B = 11
9B + C = 66
9C + D = 25
D = -2
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Subtract fourth equation from third equation. Then divide fourth equation by 9. | |
A + B = 11
9B + C = 66
C = 3
D = -2
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Subtract third equation from second equation. Then divide the resulting second equation by 9. | |
A + B = 11
B = 7
C = 3
D = -2
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Subtract the second equation from the first equation. | |
A = 4
B = 7
C = 3
D = -2
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Done. | |
Step 5: Put the solved coefficients in and integrate. Go back to your expansion into partial fractions (step 1) and substitute A, B, C, and D with the solutions you just got. Then integrate the result. That last part is pretty easy. Then click here.
Here it is
11x3 - 107x + 108 |
4 7 |
3
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2
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dx = |
3
4 ln|x + 7| + 7 ln|x - 2| - |
1
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