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# Partial fractions in integration

In integral calculus, the use of partial fractions is required to integrate the general rational function.

The integral of the polynomial function of the nth degree

f(x)=a0 + a1 x + a2 x2 + ... + an xn

is, by linearity of integration and by Formula 1) in Table of integrals:

F(x) = a0 x + a1 x2/2 + ... + an xn+1/(n+1) + C

A rational function f(x) = p(x)/q(x), p being of the mth and q of the nth degree, when m ≥ n can always be resolved into an polynomial function and a proper fractional function i.e. a function in which the degree of the numerator is at most n - 1.

This only requires the division by the denominator q(x) to be carried out until the order of the remainder becomes less than n. As the integration of the integer function has been given already, we have only to determine the integral of the form:

∫ p(x)/q(x) dx

in which p is of lower degree than q and they have no common root. Such an integral can sometimes be determined by using the natural logarithm integral condition.

This proper fractional function can be resolved into a sum of fractions with constant numerators and with denominators that are linear functions or powers of linear functions.

Let x1, x2, ..., xn the n rootss of q(x) be real or complex, but first let them all be different.

The proper fractional function can be resolved in only one way into partial fractions of the form:

For multiplying both sides of the equation by q(x), we have:
Substituting for q(x) everywhere its value as the product (x - x1) (x - x2) ... (x - xn) each denominator cancels; and if we put in for x any root xk of q, all terms having the factor (x - xk) disappear, leaving the term with the coefficient Ak; so that:
thus
q'(xk) cannot vanish, since q(x) = 0 has only distinct roots. Obviously, since q'(x) would vanish for a multiple root of q(x) = 0, this method does not apply when there are multiple roots. Accordingly:
The final formula is

Please note that if some of the roots are complex, one has to choose the same branch of the complex logarithm function wherever ln appears.

If q has multiple roots one can apply the following approach:

Assume q to be normal, that is, the leading coefficient is 1.

If where are different irreducible normal polynomials of degree , that is, polynomials of degree 2 with two conjugate complex roots (if q is real) and/or polynomials of degree 1 with exponents , we can write:

where the are polynomials of degree - 1 that can be found e.g. by a method similiar to the one shown above or by multiplying with the common denominator and comparing coefficients.

Thus we only need to consider the integral

If is of degree one, so is a constant and , then we have

If is of degree one, so is a constant and , then we have

If if of degree two, one can make a substitution of the form to reduce this case to . Remember that has to be positive, since otherwise would not be irreducible. Substituting reduces this to

If the result is

If one needs to consider the integral

,

which has a more complicated general solution, and is still to be covered.

Note that one can always factorize the polynomial into linear polynomials with complex coefficients, so the case where is of degree one is already covers all possible cases. It follows that can be expressed by complex logarithms.