1. MTH1170 Integration by Partial Fractions

2. Preliminary
Partial fraction decomposition is an algebraic process where a rational expression involving polynomials can be “decomposed” into a more simple form. This more simple form will be easier to integrate.

3. Preliminary
Partial fraction decomposition is helpful when we want to solve integrals of the form: As-long-as the degree of the polynomial in the numerator is lower than the degree of the polynomial in the denominator, we can decompose the expression into partial fractions.

4. How to use PFD
Ensure that the degree of the polynomial in the numerator is less than the degree of the polynomial in the denominator by using polynomial long division if required.
Break the polynomial in the denominator into its factors. Remember that an nth order polynomial will have n factors.
Create a partial fraction with every factor by using the factors as a denominator, and an unknown constant as the numerator. If a factor is repeated accommodations need to be made.
Begin to solve for the coefficients by equating the partial fractions to the original rational expression, and then using cross multiplication to combine the partial fractions.
Simplify the expression.
Because the simplified expression is true for any x, we can pick x values that help us determine the coefficients.

5. Example
Solve the following integral:

6. Example
In this case the degree of the numerator is already lower than the degree of the denominator so we can proceed to factoring the denominator.

7. Example
Now that we have the denominator factored, we can create a partial fraction with every factor by using the factors as a denominator, and an unknown constant as the numerator.

8. Example
Next we begin to solve for the coefficients by equating the partial fractions to the original rational expression, and then using cross multiplication to combine the right side of the equation. We then simplify as much as possible.

9. Example
Next, we pick x values to substitute into the equation that help simplify the process of solving for A, B, and C.

10. Example

11. Example

12. Example
Now that we have determined the coefficients, we can replace the unknowns in our original partial fractions.

13. Example
Now that we have converted the ratio of polynomials into partial fractions, the integral is much easier to solve.

14. Example – Repeated Factors
If the polynomial in the denominator has a repeated fraction, we need to account for it by forming fractions containing increasing powers of the factor in the denominator. Solve the following integral:

15. Example – Repeated Factors
Because the denominator of the rational expression has a repeated factor, we need to take care of it like this:

16. Example – Repeated Factors
Next we cross multiply the right hand side of the equation and simplify:

17. Example – Repeated Factors
Now we are tasked with finding the coefficients A, B, C, and D. We will plug in values of x that make this task easier. The two zeroing numbers are x = 0, and x = 1 so we will start with these:

18. Example – Repeated Factors

19. Example – Repeated Factors
Using these two values of x we were able to find the values for A, and D. We still need to find values for B and C. To do this we could plug in any value of x and get the same answer. I will pick small values of x to keep things simple:

20. Example – Repeated Factors

21. Example – Repeated Factors

22. Example – Repeated Factors
By using these values of x I have reduced our problem to a system of two equations. We need to solve this system by eliminating one of the variables. I choose to eliminate the C coefficient first and solve for B. Once I know the B coefficient I will plug that back in and solve for C:

23. Example – Repeated Factors
I have now found all of the required coefficients. We can re-write our partial fractions with the correct values:

24. Example – Repeated Factors
The goal was to solve a difficult integral. Now that we have decomposed the polynomial ratio into partial fractions this task will be much easier.