Heaviside Method Solution to Problem 11.7-1

This page applies the Heaviside method of partial fractions to find the solution. To see the standard method applied to this same problem, click here.

Even though using the Heaviside method is easier than the standard method, using a calculator is still recommended for this problem.

The problem is to find the indefinite integral

-6x³ + 71x² + 283x - 1614 dx (x - 4)(x - 3)(x + 5)(x + 6)

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, and all the roots are real. So to set it up, just put one of the unknowns over each binomial and equate the sum of those partial fractions to the integrand. Write that out, and then click here.

You should have

-6x³ + 71x² + 283x - 1614 = (x - 4)(x - 3)(x + 5)(x + 6)

A B C D + + + x - 4 x - 3 x + 5 x + 6

although any ordering of A, B, C, and D over the four binomials listed would be ok.

Step 2: Solve for A by multiplying the above through by (x - 4) and evaluating at x = 4. Work it through and then click here.

You should have

-6x³ + 71x² + 283x - 1614 = (x - 3)(x + 5)(x + 6)

B(x - 4) C(x - 4) D(x - 4) A + + + x - 3 x + 5 x + 6

= x = 4

-384 + 1136 + 1132 - 1614 = 3 1 × 9 × 10

Observe that the B, C, and D expressions are all eliminated because you multiplied them by zero. Hence you find that A = 3 from this equation alone.

Step 3: Solve for B by multiplying the setup through by (x - 3) and evaluating at x = 3. Work it through just like the last step, and then click here.

You should have

-6x³ + 71x² + 283x - 1614 = (x - 4)(x + 5)(x + 6)

A(x - 3) + B + x - 4

C(x - 3) D(x - 3) + x + 5 x + 6

= x = 3

-162 + 639 + 849 - 1614 = 4 (-1) × 8 × 9

Observe that the A, C, and D expressions are all eliminated because you multiplied them by zero. Hence you find that B = 4 from this equation alone.

Step 4: Solve for C by multiplying the setup through by (x + 5) and evaluating at x = -5. Work it through just like the last step, and then click here.

You should get

-6x³ + 71x² + 283x - 1614 = (x - 4)(x - 3)(x + 6)

A(x + 5) B(x + 5) + + C + x - 4 x - 3

D(x + 5) x + 6

= x = -5

750 + 1775 - 1415 - 1614 = -7 (-9) × (-8) × 1

Observe that the A, B, and D expressions are all eliminated because you multiplied them by zero. Hence you find that C = -7 from this equation alone.

Step 5: Solve for D by multiplying the setup through by x + 6 and evaluating at x = -6. Work it through just like the last step, and then click here.

You should get

-6x³ + 71x² + 283x - 1614 = (x - 4)(x - 3)(x + 5)

A(x + 6) B(x + 6) C(x + 6) + + + D x - 4 x - 3 x + 5

= x = -6

1296 + 2556 - 1698 - 1614 = -6 (-10) × (-9) × (-1)

Observe that the A, B, and C expressions are all eliminated because you multiplied them by zero. Hence you find that D = -6 from this equation alone.

Step 6: 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

-6x³ + 71x² + 283x - 1614 dx = (x - 4)(x - 3)(x + 5)(x + 6)

3 4 7 + - - x - 4 x - 3 x + 5

6 x + 6

dx =

              3 ln|x - 4|  +  4 ln|x - 3|  -  7 ln|x + 5|  -  6 ln|x + 6|  +  K

This was a whole lot easier than doing by the standard method, eh?

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