A rational function with a factorable denominator splits into simpler fractions, each easy to integrate.
Step 1: Let's Learn
Read it, or press Listen and follow the words.
Why it helps
Each piece integrates to a logarithm. The original combined fraction has no such simple antiderivative.
Setting it up
Factor the denominator, then write one fraction per distinct linear factor with an unknown constant on top.
Finding the constants
Multiply through by the denominator and substitute the values that make each factor zero. Each substitution isolates one constant.
The degree condition
The numerator degree must be lower than the denominator. If not, divide first and handle the polynomial part separately.
Repeated factors
A repeated factor needs one term per power, from the first up to the multiplicity.
Splitting a rational function
A rational function with a factorable denominator can be written as a sum of simpler fractions, each easy to integrate. The technique converts one hard integral into several easy ones.
Divide first if the top degree is too high
Partial fractions requires the numerator degree to be lower than the denominator's. If it is not, do polynomial division first and apply the technique to the remainder.
Finding the constants
Multiply through by the denominator and either substitute the roots to isolate each constant, or equate coefficients. Substituting roots is faster; equating coefficients is more general.
The result is usually logarithms
Each simple fraction integrates to a logarithm, so partial fraction integrals typically produce sums of logs. Recognising that in advance tells you whether the technique is likely to be the right one.
Step 2: Try It Yourself
Tap and try it out.
- Point(2, 0.50)
Step 3: Watch an Example
One step at a time.
Watch Diego Find the Constants
Diego decomposes 1 ÷ ((x − 1)(x − 2)).
- Step 1
He writes it as A ÷ (x − 1) plus B ÷ (x − 2).
Step 4: Your Turn
Practice makes it stick.
The Constant
Problem 1 of 2
For 1 = A(x − 2) + B(x − 1), substituting x = 2 gives B. What is B?
The Other One
Problem 2 of 2
Substituting x = 1 in the same equation gives A. What is A?
Split the Fraction
1 of 8
A denominator with 2 distinct linear factors. How many partial fractions?
2 of 8
A denominator with 3 distinct linear factors. How many partial fractions?
3 of 8
A factor repeated twice. How many terms does it need?
4 of 8
Numerator degree 3, denominator degree 2. Divide first? 1 yes, 0 no.
5 of 8
Each simple partial fraction integrates to what? 1 a logarithm, 2 a polynomial.
6 of 8
x² − 4 factors into how many linear factors?
7 of 8
Put the partial fractions process in order.
- 1Factor the denominator.
- 2Write one term per factor with unknown constants.
- 3Substitute the roots to find each constant.
- 4Check the fraction is proper, dividing first if not.
8 of 8
A factor repeated three times. How many terms?
Step 5: Quick Check
Show what you know.
Question 1 of 2
A denominator with 4 distinct linear factors. How many partial fractions?
Question 2 of 2
What must be checked before decomposing?
What You Learned
- 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.