Learning Outcomes

Upon completing this lesson, you will be able to:

• Decompose a rational function into a sum of simpler partial fractions.
• Integrate rational functions by applying the method of partial fractions for cases with linear and quadratic factors.
• Understand the utility of reduction formulas for solving integrals with integer powers of functions.
• Apply reduction formulas to solve complex trigonometric integrals.

Topics Covered

• Integration using Partial Fractions
• Reduction Formulas Revisited

I. Integration using Partial Fractions

We know how to integrate simple rational functions like $\frac{1}{x}$ or $\frac{1}{1+x^2}$. However, integrating more complex rational functions, such as $\frac{5x-3}{x^2-2x-3}$, requires a specific algebraic technique. The Method of Partial Fractions is a procedure for decomposing a complex rational function into a sum of simpler fractions that we already know how to integrate.

This method applies when the integrand is a rational function $f(x) = \frac{P(x)}{Q(x)}$ where the degree of the numerator $P(x)$ is less than the degree of the denominator $Q(x)$. (If the degree of the numerator is greater than or equal to the denominator, you must first perform polynomial long division).

The decomposition depends on the factors of the denominator $Q(x)$.

A. Case 1: Denominator with Distinct Linear Factors

If the denominator can be factored into distinct linear factors, $Q(x) = (ax+b)(cx+d)\dots$, then the partial fraction decomposition will have a term for each factor:

$$ \frac{P(x)}{Q(x)} = \frac{A}{ax+b} + \frac{B}{cx+d} + \dots $$

where $A, B, \dots$ are constants that we need to find.

Example 1

Evaluate

$$ \int \frac{5x-3}{x^2-2x-3} ,dx $$

1. Factor the denominator: $x^2-2x-3 = (x-3)(x+1)$.

2. Set up the partial fraction decomposition:

   $$ \frac{5x-3}{(x-3)(x+1)} = \frac{A}{x-3} + \frac{B}{x+1} $$

3. Solve for the constants $A$ and $B$: Multiply both sides by the common denominator $(x-3)(x+1)$:

   $$ 5x-3 = A(x+1) + B(x-3) $$

   This equation must be true for all values of $x$. We can solve for $A$ and $B$ by choosing convenient values for $x$:

   • Let $x=3$: $5(3)-3 = A(3+1) + B(3-3) \Rightarrow 12 = 4A \Rightarrow A=3$.
   • Let $x=-1$: $5(-1)-3 = A(-1+1) + B(-1-3) \Rightarrow -8 = -4B \Rightarrow B=2$.

4. Rewrite the integral:

   $$ \int \left(\frac{3}{x-3} + \frac{2}{x+1}\right) ,dx $$

5. Integrate term by term:

   $$ 3\int \frac{1}{x-3} ,dx + 2\int \frac{1}{x+1} ,dx = 3\ln|x-3| + 2\ln|x+1| + C $$

B. Other Cases

The setup for the decomposition changes for other types of factors: