Integration using Partial Fractions

Section 4.4 Integration using Partial Fractions

The previous three sections are all the tools for integrating rational functions. Now I can put them all together into a process.

If the rational function is not proper, use long division to make a proper fractions. The pieces without denominator that result from long division can be easily integrated as just polynomials.
Factor the denominator of the polynomial.
Use partial fractions to break up the proper rational function into pieces.
Use the standard four types from Section 4.1 to integrate the pieces produced by the partial fraction decomposition.

I’ll run through some examples, using the partial fraction examples from Section 4.3 and completing the integration.

Example 4.4.1.

I’ll return to the rational function from example Example 4.3.1.

\begin{equation*} \frac{x+2}{(x+1)(x-6)} = \frac{\frac{8}{7}}{x-6} + \frac{\frac{-1}{7}}{x+1} \end{equation*}

Now I integrate the two pieces. For each piece on the right, I can pull out the constants to integrate the typical linear form.

\begin{align*} \int \frac{x+2}{(x+1)(x-6)} dx \amp = \int \frac{\frac{8}{7}}{x-6} dx + \int \frac{\frac{-1}{7}}{x+1} dx\\ \amp = \frac{8}{7} \int \frac{1}{x-6} dx - \frac{1}{7} \int \frac{1}{x+1} dx\\ \amp = \frac{8}{7} \ln |x-6| - \frac{1}{7} \ln |x+1| + c \end{align*}

Example 4.4.2.

I’ll return to the rational function from example Example 4.3.2.

\begin{equation*} \frac{x^2+1}{(x+1)(x-2)(x-3)} = \frac{\frac{14}{9}}{x+1} - \frac{\frac{25}{9}}{x-2} + \frac{\frac{10}{3}}{x-3} \end{equation*}

Now the integral breaks up into three integrals. I can pull out the constants, so I get three ordinary linear integrals.

\begin{align*} \int \frac{x^2+1}{(x+1)(x-2)(x-3)} dx \amp = \int \frac{\frac{14}{9}}{x+1} dx - \int \frac{\frac{25}{9}}{x-2} dx + \int \frac{\frac{10}{3}}{x-3} dx\\ \amp = \frac{14}{9} \int \frac{1}{x+1} dx - \frac{25}{9} \int \frac{1}{x-2} dx + \frac{10}{3} \int \frac{1}{x-3} dx\\ \amp = \frac{14 \ln |x+1|}{9} - \frac{25\ln |x-2|}{9} + \frac{10 \ln |x-3|}{3} + c \end{align*}

Example 4.4.3.

I’ll return to the rational function from example Example 4.3.3.

\begin{equation*} \frac{x^2-x-1}{(x-1)(x^2-2x+3)} = \frac{\frac{-1}{2}}{x-1} + \frac{\frac{3}{2} x - \frac{1}{2}}{x^2 - 2x +3} \end{equation*}

Then I integrate. The linear form is familiar and easy by thie point. For the quadratic form, I have to do a bunch more work. There were two basic form in Section 4.1 and I’m going to need them both. I’m going to try to use the substitution form first, where the numerator looks like the derivative of the denominator. I need to carefully work with the algebra to set this up. Let me just focus on this quadratic piece alone. There are some algebraic tricks here that might be new: keep in mind that I’m doing whatever I need to do (adding, multiplying) to construct the desired form for the rational function.

\begin{equation*} \int \frac{\frac{3}{2} x - \frac{1}{2}}{x^2 - 2x +3} dx \end{equation*}

The derivative of the denominator is \(2x - 2\text{.}\) I want to get this expression in the numerator. To do this, I write \(\frac{3}{2} x = \frac{3}{4} 2x\text{.}\) Then I make a similar adjustment to factor \(\frac{3}{4}\) out of the whole expression.

\begin{equation*} = \int \frac{ \frac{3}{4} 2x - \frac{1}{2}}{x^2-2x+3} dx = \int \frac{ \frac{3}{4} 2x - \frac{3}{4} \frac{2}{3}}{x^2-2x+3} dx = \frac{3}{4} \int \frac{2x - \frac{2}{3}}{x^2-2x+3} dx \end{equation*}

Now the \(2x\) is correct, but the constant term in the numerator isn’t. It needs to be \(-2\text{.}\) To make this happen, I add and subtract \(-2\) in the numerator.

\begin{equation*} = \frac{3}{4} \int \frac{2x + (- 2 + 2) - \frac{2}{3})} {x^2-2x+3} dx = \frac{3}{4} \int \frac{(2x - 2) + \frac{4}{3}} {x^2-2x+3} dx \end{equation*}

Then I split the integral up into two pieces. (Not in the last step, I can pull out the \(\frac{4}{3}\) from the numerator, which cancels with the existing fraction outside the integral).

\begin{align*} \frac{3}{4} \int \frac{(2x - 2) - \frac{4}{3}} {x^2-2x+3} dx \amp = \frac{3}{4} \int \frac{(2x - 2)} {x^2-2x+3} dx + \frac{3}{4} \int \frac{\frac{4}{3}} {x^2-2x+3} dx\\ \amp = \frac{3}{4} \int \frac{(2x - 2)} {x^2-2x+3} dx + \int \frac{1} {x^2-2x+3} dx \end{align*}

The first integral now has the correct form, so I can finish that piece. For the second integral, I need to complete the square.

\begin{equation*} = \frac{3}{4} \ln |x^2-2x+3| + \int \frac{1}{(x-1)^2 +2} dx \end{equation*}

Now this second integral has the correct form as well for the arctangent integral, with \(\alpha - 1\) and \(\beta = \sqrt{2}\text{.}\) (Recall that the constant in the denominator is \(\beta^2\text{,}\) so since the constant is \(2\text{,}\) \(\beta\) must be \(\sqrt{2}\text{.}\)

\begin{align*} \amp = \frac{3}{4} \ln |x^2-2x+3| + \int \frac{1}{(x-1)^2 +2} dx \\ \amp = \frac{3}{4} \ln |x^2-2x+3| + \frac{1}{\sqrt{2}} \arctan \left( \frac{x-1}{\sqrt{2}} \right) \end{align*}

This finishes the quadratic piece. I can now write the entire integral by also doing the familiar linear piece.

\begin{align*} \amp \int \frac{x^2-x-1}{(x-1)(x^2-2x+3)} dx = \frac{-1}{2} \int \frac{1}{x-1} dx + \int \frac{\frac{3}{2} x + \frac{1}{2}}{x^2 - 2x +3} dx\\ \amp = \frac{-1}{2} \ln |x-1| + \frac{3}{4} \ln |x^2-2x+3| + \frac{1}{\sqrt{2}} \arctan \left( \frac{x-1}{\sqrt{2}} \right) + c \end{align*}

Example 4.4.4.

I’ll return to the rational function from example Example 4.3.4.

\begin{equation*} \frac{2x-3}{(x-1)(x+2)^2} = \frac{\frac{-1}{9} }{x-1} + \frac{\frac{1}{9}}{x+2} + \frac{\frac{7}{3}}{(x+2)^2} \end{equation*}

The integral splits up into three integrals. After pulling out the constants, these directly use the two linear forms from the four basic types of rational integral.

\begin{align*} \int \frac{2x-3}{(x-1)(x+2)^2}dx \amp = \frac{-1}{9} \int \frac{1}{x-1} dx + \frac{1}{9} \int \frac{1}{x+2} dx + \frac{7}{3} \int \frac{1}{(x+2)^2}dx\\ \amp = \frac{-1}{9} \ln |x-1| + \frac{1}{9} \ln |x+2| + \frac{7}{3} \frac{-1}{x+2} + c\\ \amp = \frac{-1}{9} \ln |x-1| + \frac{1}{9} \ln |x+2| - \frac{7}{3(x+2)} + c \end{align*}