মূলদ ফাংশনের যোগজীকরণ

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Some rational functions are easy to integrate. Take the integral ∫dxx−r\int \frac{dx}{x-r} for example. Recall that ∫duu=ln⁡∣u∣+C\int \frac{du}{u} = \ln |u| + C. In this case, the integral becomes ln⁡∣x−r∣+C\ln |x-r| + C. Most rational functions, however, are much more complex than this. Wouldn't it be convenient if all rational functions could be converted into a sum of these and other easy-to-integrate forms? As is turns out, they can!

It might be helpful to think back to even and odd functions to refresh your memory on how to do this.

A general rational function is of the form f(x)g(x)\frac{f(x)}{g(x)}, where ff and gg are polynomials. we have two main cases, deg⁡f≥deg⁡g\deg f \geq \deg g and deg⁡f<deg⁡g\deg f < \deg g. As it turns out, the first case can easily be reduced to the second case. How might we do that?

If we use polynomial division on the fraction, we get f(x)g(x)=q(x)+r(x)g(x)\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}, where deg⁡r<deg⁡g\deg r < \deg g. The polynomial q(x)q(x) is easily integrable, but how can we integrate the rational function r(x)g(x)\frac{r(x)}{g(x)}? We must first decompose it using partial fractions.

Let g(x)=(x−r1)(x−r2)...(x−rn)g(x) = (x-r_1)(x-r_2)...(x-r_n), i.e. an nnth-degree polynomial with no repeated roots. (We will sort out the case of repeated roots later.) We seek to find A,B,...A, B, ... such that

Ax−r1+Bx−r2+...=f(x)g(x)\frac{A}{x-r_1} + \frac{B}{x-r_2} + ... = \frac{f(x)}{g(x)}.

Placing all these partial fractions under a common denominator (i.e., g(x)g(x)) and setting coefficients equal for the two numerators creates an nn-variable system of nn equations, which we can solve for the coefficients, resulting in a sum of fractions we already know how to integrate!

In the case of repeated roots, simply add more fractions with higher power denominators. For example, if root kk appears 3 times, the sum would include Ax−rk+B(x−rk)2+C(x−rk)3\frac{A}{x-r_k} + \frac{B}{(x-r_k)^2} + \frac{C}{(x-r_k)^3}.

Let's try an example.

What would be the partial fraction decomposition of 5x2+28x+24(x−2)(x+3)(x+3)\frac{5x^2 + 28x + 24}{(x-2)(x+3)(x+3)}?

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