Integration by Partial Fractions
Integration by Partial Fractions
Partial fraction decomposition can help you with differential equations of the following form:
In solving this equation, we obtain
The problem is that we have no technique for evaluating the integral on the left side.A technique called integration by partial fractions, in its broadest applications handles a variety of integrals of the form
where p and q are polynomial functions. The technique of partial fractions becomes more complicated as the polynomials become more complicated. We shall illustrate the technique via some examples of special cases.
Example:
Solution Note that the denominator of the integrand can be factored:
The plan is to decompose this fraction into partial fractions by finding numbers A and B for which
holds for all x except x = 1 and x = - 2. If this is possible, then we can integrate 1/(x^2+x-2) by finding :
since these last two antiderivatives can be evaluated easily in terms of the natural logarithm.
We shall now show how to find A and B. Note that if we multiply both sides of the equation
by (x - l)*(x + 2), we obtain1 = A (x + 2) + B (x - 1).The last equation must hold for all x, that is, it is an identity. Since it holds for all x, it must hold for any specific values of x that we choose. Observe that if we choose x = - 2, then the term involving A will become 0, and we havel = A(-2+2)+B(-2-1)= -3Bfrom which we immediately get B = -1/3 . If we next choose x = 1, we have1 = A (1+2)+B(1-1) = 3A,and consequently A = 1/3 . Substituting these values of A and B into Formula (2), we obtain
Thus, we use partial fractions to express the fraction on the left in Equation (2).
We can now complete the integration problem.
In solving this equation, we obtain
The problem is that we have no technique for evaluating the integral on the left side.A technique called integration by partial fractions, in its broadest applications handles a variety of integrals of the form
where p and q are polynomial functions. The technique of partial fractions becomes more complicated as the polynomials become more complicated. We shall illustrate the technique via some examples of special cases.
Example:
Solution Note that the denominator of the integrand can be factored:
The plan is to decompose this fraction into partial fractions by finding numbers A and B for which
holds for all x except x = 1 and x = - 2. If this is possible, then we can integrate 1/(x^2+x-2) by finding :
since these last two antiderivatives can be evaluated easily in terms of the natural logarithm.
We shall now show how to find A and B. Note that if we multiply both sides of the equation
by (x - l)*(x + 2), we obtain1 = A (x + 2) + B (x - 1).The last equation must hold for all x, that is, it is an identity. Since it holds for all x, it must hold for any specific values of x that we choose. Observe that if we choose x = - 2, then the term involving A will become 0, and we havel = A(-2+2)+B(-2-1)= -3Bfrom which we immediately get B = -1/3 . If we next choose x = 1, we have1 = A (1+2)+B(1-1) = 3A,and consequently A = 1/3 . Substituting these values of A and B into Formula (2), we obtain
Thus, we use partial fractions to express the fraction on the left in Equation (2).
We can now complete the integration problem.