The integral of the polynomial function of the n^{th} degree

- f(
*x*)=*a*_{0}+*a*_{1}*x*+*a*_{2}*x*^{2}+ ... +*a*_{n}*x*^{n}

- F(
*x*) =*a*_{0}*x*+*a*_{1}*x*^{2}/2 + ... +*a*_{n}*x*^{n+1}/(*n*+1) +*C*

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*

*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.*

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

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.