P3 Algebra: Partial Fractions

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1. The Concept

Partial fractions allow us to split complex fractions into simpler ones. This is crucial for integration.

Interactive Check:

If we want to split $$ \frac{5x-1}{(x-1)(x+3)} $$ into partial fractions, what should the structure look like?

2. Solving Linear Factors

Let's solve for the constants $A$ and $B$ together.

Problem: $$ \frac{5x+7}{(x+1)(x+2)} \equiv \frac{A}{x+1} + \frac{B}{x+2} $$

Step 1: Remove Denominators

Multiply everything by $(x+1)(x+2)$. What is the resulting equation?

$$ 5x+7 \equiv A(\dots) + B(\dots) $$

$A$ is multiplied by: $B$ is multiplied by:

Tip: Type "x+2" and "x+1". Press Enter or Click outside the box to check.
Step 2: Substitution Method

We have: $5x+7 = A(x+2) + B(x+1)$

To find A, let $x = -1$. This eliminates $B$.

Left Side ($5(-1)+7$): 2

Right Side ($A(-1+2)$): 1A

So, $A =$

Step 3: Find B

To find B, let $x = -2$. This eliminates $A$.

Left Side ($5(-2)+7$):

Right Side ($B(-2+1)$): $-1B$

So, $-3 = -B \implies B =$

🎉 Solved!

$$ \frac{5x+7}{(x+1)(x+2)} \equiv \frac{2}{x+1} + \frac{3}{x+2} $$

3. The Repeated Factor Trap

When you have $(x+3)^2$, you need a term for power 1 AND power 2.

Task: Build the correct form for: $$ \frac{9}{(x-1)(x+2)^2} $$

Term 1
+
Term 2
+
Term 3

Correct! The identity is:

$$ 9 \equiv A(x+2)^2 + B(x-1)(x+2) + C(x-1) $$

If we let $x=-2$, we can find $C$ immediately.

$9 = C(-2-1) \implies 9 = -3C$

$C =$

4. Improper Fractions (Top Heavy)

If degree of Top $\ge$ Bottom, you MUST divide first.

$$ \text{Example: } \frac{x^2 + 1}{x^2 - 1} $$

Step 1: Check Degrees
Numerator: $x^2$ (Degree 2). Denominator: $x^2$ (Degree 2).
$2 \ge 2$, so it is Improper.
Step 2: Long Division
How many times does $x^2$ go into $x^2$? 1 time.
$$ (x^2+1) \div (x^2-1) = 1 \text{ rem } 2 $$
Step 3: Rewrite
$$ 1 + \frac{2}{x^2-1} $$
Step 4: Solve Partial Fractions for Remainder
$$ \frac{2}{(x-1)(x+1)} = \frac{1}{x-1} - \frac{1}{x+1} $$
Final Answer: $$ 1 + \frac{1}{x-1} - \frac{1}{x+1} $$