Partial Fractions

Splitting a rational function into pieces.

12 problems · about 22 minutes · FUN-6.F

What this lesson teaches

The student decomposes rational functions into partial fractions and integrates them.

Partial fractions split a rational function into pieces that integrate to logarithms.
Substituting each root isolates one constant at a time.
Divide first if the numerator degree is not lower.

Straightforward practice. Get the method working first.

5 problems

A denominator with 4 distinct linear factors. How many partial fractions?
Answer 4
Why 4.
What must be checked before decomposing?
Answer That the numerator degree is lower than the denominator.
Why The fraction must be proper.
A denominator with 2 distinct linear factors. How many partial fractions?
Answer 2
Why One per factor.
A denominator with 3 distinct linear factors. How many partial fractions?
Answer 3
Why One per factor.
A factor repeated twice. How many terms does it need?
Answer 2
Why One per power.

The same ideas with more to keep track of.

3 problems

Numerator degree 3, denominator degree 2. Divide first? 1 yes, 0 no.
Answer 1
Why The fraction is not proper.
Each simple partial fraction integrates to what? 1 a logarithm, 2 a polynomial.
Answer 1
Why One over a linear factor.
x² − 4 factors into how many linear factors?
Answer 2
Why (x − 2)(x + 2).

Mixed problems. Work out what kind of question it is before you start.

4 problems

Put the partial fractions process in order.
Answer 1. Check the fraction is proper, dividing first if not. 2. Factor the denominator. 3. Write one term per factor with unknown constants. 4. Substitute the roots to find each constant.
Why The degree check comes before anything else.
A factor repeated three times. How many terms?
Answer 3
Why One per power.
The Constant: For 1 = A(x − 2) + B(x − 1), substituting x = 2 gives B. What is B?
Answer 1
Why 1.
The Other One: Substituting x = 1 in the same equation gives A. What is A?
Answer -1
Why −1.