Welcome back! Today is our last day of adding integration techniques to our toolbox. We saw this tool in action in Lesson 8 (link here) and today we’ll see how to apply it in a wider variety of examples.

Lesson 9: Partial Fraction Decomposition (part 2)

#### Learning goals:

- Integrate a rational function using the method of partial fractions.
- Recognize quadratic factors in a rational function.

**Topic**:

- Volume 2, Section 3.4
*Partial Fraction Decomposition*(link to textbook section)

**WeBWorK**:

- Integration – Partial Fractions

**Motivating question**

How can we evaluate \[\int \frac{x^2+2x-1}{x^3+x^2+x+1}dx?\]

Evaluate:

\[\int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx\]

Just like in Warmup exercise 1 from Lesson 8 (link here), we are lucky because we can integrate by substitution. Let $u= x^3+x^2+x+1$ so $du = (3x^2+2x+1)dx$. Then

\begin{align} \int \frac{3x^2+2x+1}{x^3+x^2+x+1}dx &= \int \frac{du}{u} \\ &=\ln(|u|) + C\\ & = \ln(|x^3+x^2+x+1|)+C \end{align}

Evaluate:

\[\int \left(\frac{2x}{x^2+1} + \frac{1}{x+1} \right)dx\]

Just like in Warmup exercise 2 from Lesson 8 (link here), we can integrate term-by-term. We’ll use substitution to integrate the first term, with $u=x^2+1$ so $du = 2x dx$.

\begin{align} \int \left(\frac{2x}{x^2+1} + \frac{1}{x+1} \right)dx &=\int \frac{du}{u} + \int \frac{1}{x+1}dx \\ &= \ln(|u|) + \ln(|x+1|)+ C\\ &= \ln(|x^2+1|)+ \ln(|x+1|)+ C \end{align}

Subtract: \[\frac{2x}{x^2+1} + \frac{1}{x+1}\]

Just like in Warmup exercise2 from Lesson 8 (link here), we need to find the lowest common denominator. In this case, the LCD is $(x^2+1)(x-1)$. So:

\begin{align} \frac{2x}{x^2+1} + \frac{1}{x+1} &= \frac{2x(x+1) – (x^2+1)}{(x^2+1)(x-1)}\\ &= \frac{2x^2+2x-x^2-1}{x^3+x^2+x+1} \\ &= \frac{x^2+2x-1}{x^3+x^2+x+1} \end{align}

Just like in Lesson 8, Warmup exercise 3 is the key to unlocking the motivating example!

#### Motivating example

\begin{align} \int \frac{x^2+2x-1}{x^3+x^2+x+1}dx &\stackrel{{\textrm{ how tho?}}}{=}\int \left(\frac{2x}{x^2+1} + \frac{1}{x+1} \right)dx \\ &= \ln(|x^2+1|)+ \ln(|x+1|)+ C \end{align}

## Fundamental theorem of algebra

Again, partial fraction decomposition is an algebra technique, not a calculus technique. Remember the version of the fundamental theorem of algebra that we want:

**Theorem:** Any polynomial with real coefficients can be factored uniquely into linear and irreducible factors.

#### Header text

In Lesson 8, we saw how linear factors in the denominator contribute to the partial fraction decomposition of a rational expression. We saw that when $(x-c)^n$ appears in the factored denominator, then there are terms with denominators $x-c$, $(x-c)^2$, $\dots$, $(x-c)^n$ and constant numerators.

Today we’ll see how the irreducible quadratic factors (the quadratic factors that cannot be factored into a product of two linear factors) contribute to the partial fraction decomposition of a rational expression.

## Examples

**Video 1** below shows an example of a partial fraction decomposition where the denominator factors into one linear factor and one irreducible quadratic factor. Notice that this example didn’t even involve any integration. However, if you were asked to evaluate

\[\int \frac{7x^2+11x+46}{x^3+2x^2+9x+18}dx\]

then the Video 1 does almost all the work.

**Video 2** also shows only the algebraic part; if you had been asked to evaluate

\[\int \frac{10x^2+12x+20}{x^3-8}\]

Video 2 would do most of the work for you. Notice that long division of polynomials is used to factor the denominator in Video 2. If you need a refresher on long division of polynomials, see the lesson in the MAT 1375 course hub (link here).

One important detail to remember when finding the partial fraction decomposition of a rational expression is that the degree of the numerator must be *strictly lower* than the degree of the denominator. If the degree of the numerator is equal to or higher than the degree of the denominator, we have to apply long division of polynomials first and then find the partial fraction decomposition of the remainder term. **Video 3 **shows an example of this. Notice that in Video 3, the exercise is to evaluate an indefinite integral, so the video shows the algebra steps as well as the calculus steps.

### Additional Video Resources

Evaluate:

\[ \int \frac{4}{(x-1)(x^2+9)}dx \]