L&T, 2.12.2.1
Before dealing with partial fractions, we need to define a rational function.
Partial fractions is a method of simplifying a rational function. For the present we shall only consider rational functions where the degree of the numerator is less than that of the denominator ( not equal). If this is not true then we can convert it into this form–see later (integration section). First factorise the denominator into a mixture of linear and quadratic factors. This can always be done without using complex numbers (use linear factors only if possible). E.g.,
We can now simplify the rational function using partial fractions. We do this by means of examples as part of the revision.
Example 5.19:
Simplify using partial fractions.
Solution:
We have factorised the denominator). Now put
( constants). Multiply both sides by , the denominator of the left-hand side. We find
(5.3) |
Now compare coefficients on both sides. First : , and for the constant term we find . We solve these simultaneous linear equations, and find . So
Alternatively we can find A and B by choosing values for . If we choose then ( 5.3 ) becomes , and therefore . If we choose then it becomes , and therefore , in agreement with our previous results.
Example 5.20:
Simplify using partial fractions.
Solution:
(left as an exercise, , , ).
Example 5.21:
Simplify using partial fractions.
Solution:
where we have one term for each power of the factor up to the maximum. Multiply by and equate coefficients.
Substitute : so . : so . Equate the coefficients of : , so , and the coefficients of constant term: , and thus .
Example 5.22:
Simplify using partial fractions.
Solution:
First factorise , . We cannot factorise into their factors with real coefficients. Write
Multiply with ,
substitute : , or . Equate coefficients of : , . Equate coefficients of the constant part: , .
A rational function is a function of the form where and are both polynomials.
Integration of such functions are dealt with according to the following procedure:
(5.4) |
where and are polynomials and has lower degree than ,
Example 5.23:
Bring to the form ( 5.4 ).
Solution:
Put This corresponds to and . Thus . We can clearly integrate directly (why?).
We now obtain integrals with one or more of the following types
Example 5.24:
Integrate .
Solution:
This integrand can be rewritten as
To find , , , we need to solve
We can get one of the values for almost free, using : , or . We solve for the rest by equating the coefficients of identical powers of ,
We have reexpressed the integral as
The first term ( ) is easy to integrate and gives . Let us therefore concentrate on the second term
Here we have used the fact that the differential of the denominator is . The remaining integral is treated by completing the square,
which allows us to write
Using the two previous examples we conclude that