In this section we apply techniques from previous sections to compute integrals of rational functions over the real numbers. The new ingredient is partial fraction decomposition. This method enables us to express a rational function, ( co-prime polynomials) as the sum of a polynomial and rational functions whose denominators are powers of irreducible factors of .
A non-zero polynomial is monic if its leading coefficient is . A non-constant polynomial is irreducible if it is not a product of two polynomials of positive degree. Two polynomials are co-prime if they do not have any positive degree factor in common.
Every polynomial is a product of irreducible polynomials. Moreover, the monic irreducible factors of a polynomial are uniquely determined by the polynomial.
A linear (degree ) polynomial is clearly irreducible. A quadratic polynomial is reducible if and only if it has a linear factor. Dividing a polynomial by a linear polynomial (), we get
Therefore, is a root of if and only if divides . As a result, a quadratic polynomial is reducible over the real numbers if and only if it has a real root if and only if its discriminant, , is non-negative.
The assumptions on and in the theorem above can always be achieved: by dividing the leading coefficient of into , we can assume is monic. We should point out that requiring to be monic only makes the statement simpler but not essential. By canceling common non-constants factors of and , we can assume they are co-prime. Doing this potentially extends the domain of the rational function by finitely many points. With them taken out, the indefinite integral of and that for the reduced fraction are the same.
We illustrate how to compute partial fraction decomposition by a few examples. From the discussion above, we only need to focus on the case where the rational function is written as where and are co-prime polynomials and that . In this case, we call the denominator of . Note that is determined by only up to a non-zero multiplicative constant.
Let us find the partial fraction decomposition of the rational function . The denominator of is reducible since its discriminant , is positive. Indeed, factors as . Let us put this in the form of a table:
Table3.39.Table for the partial fractions decomposition of F(x)
According to Theorem 3.37, each irreducible factor of corresponding to a group of summands. The multiplicity of that factor is the number of summands in that group (one for each power). And the degree of that factor determines the form of the numerator in each of the summand in that group. If the degree is 1 then the corresponding numerator is simply a constant. If the degree is 2, then the corresponding numerator is of a linear form.
We can find and by comparing coefficients which transfers the problem into one about solving a system of linear equations (and hence can be handled by linear algebra in general). For simplicity, we relabel the coefficients and as and , respectively.
By comparing of the numerators, we get
(3.8)
and so,
and(3.9)
Solving this system of linear equations yields , therefore the partial fractions decomposition of is
Let us compute the partial fraction decomposition of
This time is not smaller than the degree of (both are of degree ). So, the polynomial part of the decomposition is not zero. We can find that out by dividing by , say using long division. We find that. Thus,
One checks that and have no irreducible factors in common. By uniqueness of partial fraction, it remains to compute the partial fraction decomposition of . Again we organize the information obtained from the factorization of in a table:
Table3.41.Table for the partial fractions decomposition of F(x)
The identity holds for all so instead of comparing coefficients, one can also derive linear relations among the coefficients by setting to various values. We aim at getting these relations as simple as possible, for instance, by setting , we arrive at
That is and so . Next by setting , equivalently by comparing the constant terms on both sides, we get . So, .
We can mix these techniques: at this point by comparing the leading coefficients on both sides, we get , so . Comparing the coefficients of the term leads us to
and so . At this juncture, we need one more relation among and . We can do that by substituting another value for . For example, we get by setting (checks that yourself). And so together with we conclude that .
Let us compute the partial fraction decomposition of
Note that has degree larger than and so must be reducible over the real numbers. In fact, the intermediate value theorem guarantees a real root for a cubic polynomial over and hence a linear factor. For instance, in this particular case,
so must of a root between and . Keep in mind, sketching the graph of a function often give you some insights on where are the roots located.
In general, the roots of a cubic can be expressed in terms of its coefficients by arithmetic operations and taking surds. However, we can guess the "nice" roots if there are some. Suppose we have a factorization
where is an integer. Then and hence must be a factor of . And given we know that there is a root of between and , it is easy to check that is a root of . Thus, by dividing by , we have
Since has no real roots, it is irreducible. Therefore, the partial fraction decomposition of looks like
By multiplying on both sides, we get to
Setting yields , so . Then by setting , we get
Thus, . And by comparing the coefficients of on both sides, we see that , so .
With partial fraction decomposition the problem of integrating rational functions over the real numbers reduces to the one of finding integrals of the form
In the second case, is irreducible. So by completing square and a linear change in variable, the problem boils down to finding integrals of the following two kinds
The first kind can be handled by the substitution and the second kind is handled by the trigonometric substitution . Let us illustrate these strategy by an example.