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 n^th degree

f(x)=a₀ + a₁ x + a₂ x² + ... + a_n xⁿ

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

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

A rational function f(x) = p(x)/q(x), p being of the m^th and q of the n^th 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, ..., x_n the n roots 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:

\frac{p \left( x \right)}{q \left( x \right)} = \frac{A_1}{x - x_1} + \frac{A_2}{x - x_2} + \cdots + \frac{A_n}{x - x_n}

For multiplying both sides of the equation by q(x), we have:

p \left( x \right) = \frac{A_1}{x - x_1} q \left( x \right) + \frac{A_2}{x - x_2} q \left( x \right) + \cdots + \frac{A_n}{x - x_n} q \left( x \right)

Substituting for q(x) everywhere its value as the product (x - x₁) (x - x₂) ... (x - x_n) each denominator cancels; and if we put in for x any root x_k of q, all terms having the factor (x - x_k) disappear, leaving the term with the coefficient A_k; so that:

p \left( x_k \right) = A_k \lim_{x \rightarrow x_k}{\frac{q \left( x \right) - \overbrace{q\left(x_k\right)}^{=0}}{x - x_k}} = A_k q' \left( x_k \right)

thus

A_k = \frac{p \left( x_k \right)}{q'\left( x_k \right)}

q'(x_k) 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:

\frac{p \left( x \right)}{q \left( x \right)} = \sum_{k = 1}^{n}{\frac{p \left( x_k \right)}{q'\left( x_k \right)} \frac{1}{x - x_k}}

The final formula is

\int{\frac{p \left( x \right)}{q \left( x \right)}\,dx} = \frac{p \left( x_1 \right)}{q'\left( x_1 \right)} \ln \left( x - x_1 \right) + \frac{p \left( x_2 \right)}{q'\left( x_2 \right)} \ln \left( x - x_2 \right) + \cdots + \frac{p \left( x_n \right)}{q'\left( x_n \right)} \ln \left( x - x_n \right) + C

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

q(x) = q_1(x)^{e_1}\cdot q_2(x)^{e_2}\cdots q_n(x)^{e_n}

where

q_i

are different irreducible normal polynomials of degree

\deg(q_i)

, that is, polynomials of degree 2 with two conjugate complex roots (if q is real) and/or polynomials of degree 1 with exponents

e_i

, we can write:

q(x) = \frac{p_{11}}{q_1} + \frac{p_{12}} + \frac{p_{21}}{q_2} + \ldots + \frac{p_{2e_2}} + \ldots + \frac{p_n}

where the

p_{ij}

are polynomials of degree

\deg(q_i)

- 1 that can be found e.g. by a method similar to the one shown above or by multiplying with the common denominator and comparing coefficients. Thus we only need to consider the integral

\int \frac{p_{ij}}

q_i = x^2 + Ax + B

if of degree two, one can make a substitution of the form

u:= x - \frac A2

to reduce this case to

q_i = u^2 + x_i

. Remember that

x_i

has to be positive, since otherwise

q_i

would not be irreducible. Substituting

v:=\frac u{\sqrt{x_i}}

reduces this to

x_i = 1

j = 1

the result is

\int \frac{Ax+B}{x^2 + 1} dx = \frac A2 \ln(x^2+1) + B \arctan(x) + C

j > 1

one needs to consider the integral

\int \frac{Ax+B} + A \int \frac 1{{(x^2 + 1)}^j}

, 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

q_i

is of degree one is already covers all possible cases. It follows that

\arctan

can be expressed by complex logarithms.

External link

* Partial Fraction Expander http://mss.math.vanderbilt.edu/~pscrooke/MSS/partialfract.html

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