Standard Method Solution to Problem 11.7-2

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

6x³ + 55x² + 224x + 1980 dx (x - 6)(x + 5)(x² + 4x + 40)

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. There are no repeated roots in the denominator, but only to of the roots are real (that's because that quadratic can't be factored). So to set it up, just put one of the unknowns over each binomial, and put the other two over the quadratic the way we did in the main text. Then equate the sum of those partial fractions to the integrand. Write that out, and then click here.

You should have

6x³ + 55x² + 224x + 1980 = (x - 6)(x + 5)(x² + 4x + 40)

A B Cx + D + + x - 6 x + 5 x² + 4x + 40

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 - 6)(x + 5),
(x - 6)(x² + 4x + 40), and
(x + 5)(x² + 4x + 40).
Two of these will each be multiplied by one of the unknowns. The other will be multiplied by (Cx + D). 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 + 5)(x² + 4x + 40) = x³ + 9x² + 60x + 200. You multiply B by (x - 6)(x² + 4x + 40) = x³ - 2x² + 16x - 240. You multiply (Cx + D) by (x + 5)(x - 6) = x² - x - 30 to get Cx³ + (-C + D)x² + (-30C - D)x - 30D. Take the sum of those products, gather like powers of x, and you have

6x³ + 55x² + 224x + 1980 = (x - 6)(x + 5)(x² + 4x + 40)

(A + B + C)x³ + (x - 6)(x + 5)(x² + 4x + 40)

(9A - 2B - C + D)x² + (x - 6)(x + 5)(x² + 4x + 40)

(60A + 16B - 30C - D)x + (x - 6)(x + 5)(x² + 4x + 40)

(200A - 240B + 0C - 30D) + (x - 6)(x + 5)(x² + 4x + 40)

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

6 = A + B + C
equating coefficients of the x³ terms
55 = 9A - 2B - C + D
equating coefficients of the x² terms

224 = 60A + 16B - 30C - D
equating coefficients of the x terms

1980 = 200A - 240B - 30D
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 + C = 6 9A - 2B - C + D = 55 60A + 16B - 30C - D = 224 200A - 240B - 30D = 1980
Subtract 9 times first equation from second equation. Subtract 60 times first equation from third equation. Subtract 200 times first equation from fourth equation.

A + B + C = 6 -11B - 10C + D = 1 -44B - 90C - D = -136 -440B - 200C - 30D = 780
Subtract 4 times second equation from third equation. Subtract 40 times second equation from fourth equation.

A + B + C = 6 -11B - 10C + D = 1 -50C - 5D = -140 200C - 70D = 740
Add 4 times third equation to fourth equation. Then divide fourth equation by -90

A + B + C = 6 -11B - 10C + D = 1 -50C - 5D = -140 D = -2
Add 5 times fourth equation to third equation. Then divide fourth equation by -50.

A + B + C = 6 -11B - 10C + D = 1 C = 3 D = -2
Subtract fourth equation from second equation. Add 10 times third equation to second equation. Then divide the resulting second equation by -11.

A + B + C = 6 B = -3 C = 3 D = -2
Subtract the second and third, equations both from the first.

A = 6 B = -3 C = 3 D = -2
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. Then click here.

Here it is

6x³ + 55x² + 224x + 1980 dx = (x - 6)(x + 5)(x² + 4x + 40)

6 3 - + x - 6 x + 5

3x - 2 x² + 4x + 40

dx

This one is not as easy to integrate as the first. You have complete the square on the third partial fraction. The substitution for that is

v = x + 2
and
v² = x² + 4x + 4
hence
v² + 36 = x² + 4x + 40

The derived substitutions that you also need for this are v - 2 = x and dv = dx.

6 3 - + x - 6 x + 5

3x - 2 x² + 4x + 40

dx =

6 dx - x - 6

3 dx + x + 5

3v - 8 dx v² + 36

You break the third integral into two integrals. The first yields to simple substitution. It's substitution is

s = v² + 36
and
1 ds = v dv 2

The second requires trigonometric substitution. Its substitution is

v tan(u) = 6
and
6 (1 + tan²(u)) du = dv

With these substitutions you find that

6 3 - + x - 6 x + 5

3x - 2 x² + 4x + 40

dx =

6 ln|x - 6| - 3 ln|x + 5| +

3 2

ln|x² + 4x + 40| -

4 arctan 3

x + 2 6

+ K

Note that to arrive at this, I've skipped some steps here that we have already covered in previous sections.

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6 = A + B + C		equating coefficients of the `x³` terms
55 = 9A - 2B - C + D		equating coefficients of the `x²` terms
224 = 60A + 16B - 30C - D		equating coefficients of the `x` terms
1980 = 200A - 240B - 30D		equating coefficients of the constant terms

Equations		Procedure
A + B + C = 6 9A - 2B - C + D = 55 60A + 16B - 30C - D = 224 200A - 240B - 30D = 1980		Subtract `9` times first equation from second equation. Subtract `60` times first equation from third equation. Subtract `200` times first equation from fourth equation.

A + B + C = 6 -11B - 10C + D = 1 -44B - 90C - D = -136 -440B - 200C - 30D = 780		Subtract `4` times second equation from third equation. Subtract `40` times second equation from fourth equation.

A + B + C = 6 -11B - 10C + D = 1 -50C - 5D = -140 200C - 70D = 740		Add `4` times third equation to fourth equation. Then divide fourth equation by `-90`

A + B + C = 6 -11B - 10C + D = 1 -50C - 5D = -140 D = -2		Add `5` times fourth equation to third equation. Then divide fourth equation by `-50`.

A + B + C = 6 -11B - 10C + D = 1 C = 3 D = -2		Subtract fourth equation from second equation. Add `10` times third equation to second equation. Then divide the resulting second equation by `-11`.

A + B + C = 6 B = -3 C = 3 D = -2		Subtract the second and third, equations both from the first.

A = 6 B = -3 C = 3 D = -2		Done.