Integração por funções racionais

Some rational functions are easy to integrate. Take the integral $\int \frac{dx}{x-r}$ for example. Recall that $\int \frac{du}{u} = \ln |u| + C$. In this case, the integral becomes $\ln |x-r| + C$. 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 $\frac{f(x)}{g(x)}$, where $f$ and $g$ are polynomials. we have two main cases, $\deg f \geq \deg g$ and $\deg f < \deg g$. 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 $\frac{f(x)}{g(x)} = q(x) + \frac{r(x)}{g(x)}$, where $\deg r < \deg g$. The polynomial $q(x)$ is easily integrable, but how can we integrate the rational function $\frac{r(x)}{g(x)}$? We must first decompose it using partial fractions.

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

$\frac{A}{x-r_1} + \frac{B}{x-r_2} + ... = \frac{f(x)}{g(x)}$.

Placing all these partial fractions under a common denominator (i.e., $g(x)$) and setting coefficients equal for the two numerators creates an $n$-variable system of $n$ 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 $k$ appears 3 times, the sum would include $\frac{A}{x-r_k} + \frac{B}{(x-r_k)^2} + \frac{C}{(x-r_k)^3}$.

Let's try an example.

Good job! Once you have the partial fraction decomposition, integration becomes simple. Remember to integrate fractions whose denominators have higher power than 1 through the power rule. In the case of our example,

$\int \frac{5x^2 + 28x + 24}{(x-2)(x+3)(x+3)}dx = \int \frac{4}{x-2}dx + \int \frac{1}{x+3}dx + \int \frac{3}{(x+3)^2}dx$

$= 4\ln|x-2| + \ln|x-3| - \frac{3}{x+3}$.

All rational functions can be integrated using this method.

Blurb

We can integrate any rational function using partial fraction decomposition. This method yields a sum of logarithms and arctangents.