Standard Method Solution to Problem 11.7-4

This page applies the standard method of partial fractions to find the solution. You can't do the Heaviside method on this problem (without resorting to complex-number arithmetic) because it has no real roots in its denominator.

The problem is to find the indefinite integral

3x³ + 31x² - 55x - 95 dx (x² - 6x + 25)(x² + 2x + 5)

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 one has no real roots in its denominator. The denominator consists of two irreducible quadratics. Use the same kind of setup for each of them as we used in the main text. Remember this denominator is fourth degree, so you will need four unknowns altogether. Write that out, and then click here.

You should have

3x³ + 31x² - 55x - 95 = (x² - 6x + 25)(x² + 2x + 5)

Ax + B Cx + D + x² - 6x + 25 x² + 2x + 5

although any ordering of A, B, C, and D over the four binomials listed would be ok. The ordering shown here, however, is what will be used for the duration of this solution.

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 both of the following:
(Ax + B) by (x² + 2x + 5), and
(Cx + D) by (x² - 6x + 25).
Multiply it all out, write the new partial fractions, then click here.

To get the combined numerator on the right, you multiply (Ax + B) by (x² + 2x + 5) to get Ax³ + (2A + B)x² + (5A + 2B)x + 5B. You multiply (Cx + D) by (x² - 6x + 25) to get Cx³ + (-6C + D)x² + (25C - 6D)x + 25D. Take the sum of those products, gather like powers of x, and you have

3x³ + 31x² - 55x - 95 = (x² - 6x + 25)(x² + 2x + 5)

(A + C)x³ + (x² - 6x + 25)(x² + 2x + 5)

(2A + B - 6C + D)x² + (x² - 6x + 25)(x² + 2x + 5)

(5A + 2B + 25C - 6D)x + (x² - 6x + 25)(x² + 2x + 5)

(5B + 25D) (x² - 6x + 25)(x² + 2x + 5)

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

3 = A + C
equating coefficients of the x³ terms
31 = 2A + B - 6C + D
equating coefficients of the x² terms

-55 = 5A + 2B + 25C - 6D
equating coefficients of the x terms

-95 = 5B + 25D
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 + C = 3 2A + B - 6C + D = 31 5A + 2B + 25C - 6D = -55 5B + 25D = -95
Subtract 2 times first equation from second equation. Subtract 5 times first equation from third equation.

A + C = 3 B - 8C + D = 25 2B + 20C - 6D = -70 5B + 25D = -95
Subtract 2 times second equation from third equation. Subtract 5 times second equation from fourth equation.

A + C = 3 B - 8C + D = 25 36C - 8D = -120 40C + 20D = -220
Multiply fourth equation by 9/10. Then subtract third equation from resulting fourth equation

A + C = 3 B - 8C + D = 25 36C - 8D = -120 26D = -78
Divide fourth equation by 26. Add 8 times resulting fourth equation to third equation.

A + C = 3 B - 8C + D = 25 36C = -144 D = -3
Divide third equation by 36. Add 8 times resulting third equation to second equation. Subtract fourth equation from second equation.

A + C = 3 B = -4 C = -4 D = -3
Subtract the third equation from the first equation.

A = 7 B = -4 C = -4 D = -3
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. You have to do a bunch of substitutions to get this one integrated. Do the best you can, then click here.

Here it is

3x³ + 31x² - 55x - 95 dx = (x² - 6x + 25)(x² + 2x + 5)

7x - 4 4x + 3 − x² - 6x + 25 x² + 2x + 5

dx

The right-hand integral above can of course be broken into the sum of two integrals. Each of them requires you to complete the square. Your respective substitutions are

v = x - 3
and
v² = x² - 6x + 9
hence
v² + 16 = x² - 6x + 25

and

w = x + 1
and
w² = x² + 2x + 1
hence
w² + 4 = x² + 2x + 5

You also find that v + 3 = x, dv = dx, w - 1 = x, and dw = dx. Your new integrals are

7x - 4 4x + 3 − x² - 6x + 25 x² + 2x + 5

dx =

7v + 17 dv - v² + 16

4w - 1 dw w² + 4

Each of these then breaks into two new integrals, one each needing simple substitution, the other needing trigonometric substitution. For the first one your substitutions for v are

r = v² + 16
and
1 dr = v dv 2

and

v tan(t) = 4
and
4 (1 + tan²(t)) dt = dv

For the other integral it's

s = w² + 4
and
1 ds = w dw 2

and

w tan(u) = 2
and
2 (1 + tan²(u)) du = dw

Skipping the remaining steps, which are covered in previous sections, you end up with

3x³ + 31x² - 55x - 95 dx = (x² - 6x + 25)(x² + 2x + 5)

7 2

ln|x² - 6x + 25| +

17 arctan 4

x - 3 4

-

2 ln|x² - 2x + 5| +

1 arctan 2

x + 1 2

+ K

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3 = A + C		equating coefficients of the `x³` terms
31 = 2A + B - 6C + D		equating coefficients of the `x²` terms
-55 = 5A + 2B + 25C - 6D		equating coefficients of the `x` terms
-95 = 5B + 25D		equating coefficients of the constant terms

Equations		Procedure
A + C = 3 2A + B - 6C + D = 31 5A + 2B + 25C - 6D = -55 5B + 25D = -95		Subtract `2` times first equation from second equation. Subtract `5` times first equation from third equation.

A + C = 3 B - 8C + D = 25 2B + 20C - 6D = -70 5B + 25D = -95		Subtract `2` times second equation from third equation. Subtract `5` times second equation from fourth equation.

A + C = 3 B - 8C + D = 25 36C - 8D = -120 40C + 20D = -220		Multiply fourth equation by `9/10`. Then subtract third equation from resulting fourth equation

A + C = 3 B - 8C + D = 25 36C - 8D = -120 26D = -78		Divide fourth equation by `26`. Add `8` times resulting fourth equation to third equation.

A + C = 3 B - 8C + D = 25 36C = -144 D = -3		Divide third equation by `36`. Add `8` times resulting third equation to second equation. Subtract fourth equation from second equation.

A + C = 3 B = -4 C = -4 D = -3		Subtract the third equation from the first equation.

A = 7 B = -4 C = -4 D = -3		Done.

v = x - 3	and	v² = x² - 6x + 9	hence	v² + 16 = x² - 6x + 25
and
w = x + 1	and	w² = x² + 2x + 1	hence	w² + 4 = x² + 2x + 5

r = v² + 16	and	1 dr = v dv 2
and
v tan(t) = 4	and	4 (1 + tan²(t)) dt = dv

s = w² + 4	and	1 ds = w dw 2
and
w tan(u) = 2	and	2 (1 + tan²(u)) du = dw