💡 Key Idea: Decompose the fraction into simpler fractions that we can integrate easily!

What is Partial Fraction Decomposition?

Break a complicated fraction into a sum of simpler fractions.

Example: $\frac{1}{x^2-1} = \frac{1}{(x-1)(x+1)} = \frac{A}{x-1} + \frac{B}{x+1}$

Find $A$ and $B$, then integrate each piece separately!

When to Use Partial Fractions

Requirements:

1. Rational function: $\frac{P(x)}{Q(x)}$ where $P$ and $Q$ are polynomials
2. Proper fraction: degree of $P <$ degree of $Q$
   • If not, use polynomial long division first!
3. Factored denominator: $Q(x)$ must be factored

The Four Cases

Case 1: Distinct Linear Factors

Form: $(x-a)(x-b)(x-c)\cdots$

Decomposition: $\frac{P(x)}{(x-a)(x-b)} = \frac{A}{x-a} + \frac{B}{x-b}$

Each linear factor gets one term with a constant numerator.

Case 2: Repeated Linear Factors

Form: $(x-a)^n$

Decomposition: $\frac{P(x)}{(x-a)^3} = \frac{A}{x-a} + \frac{B}{(x-a)^2} + \frac{C}{(x-a)^3}$

Include all powers from 1 up to $n$.

Case 3: Distinct Irreducible Quadratic Factors

Form: $(ax^2+bx+c)$ where $b^2-4ac < 0$ (can't factor)

Decomposition: $\frac{P(x)}{(x-a)(x^2+1)} = \frac{A}{x-a} + \frac{Bx+C}{x^2+1}$

Quadratic factor gets a linear numerator!

Case 4: Repeated Irreducible Quadratic Factors

Form: $(ax^2+bx+c)^n$

Decomposition: $\frac{P(x)}{(x^2+1)^2} = \frac{Ax+B}{x^2+1} + \frac{Cx+D}{(x^2+1)^2}$

Include all powers with linear numerators.

Example 1: Distinct Linear Factors

Integrate $\int \frac{1}{x^2-1}\,dx$

Step 1: Factor denominator

$x^2 - 1 = (x-1)(x+1)$

Step 2: Set up partial fractions

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

Step 3: Find A and B

Multiply both sides by $(x-1)(x+1)$:

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

Method 1: Substitution

Let $x = 1$: $1 = A(2) + B(0)$ → $A = \frac{1}{2}$

Let $x = -1$: $1 = A(0) + B(-2)$ → $B = -\frac{1}{2}$

Step 4: Rewrite and integrate

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

$= \frac{1}{2}\ln|x-1| - \frac{1}{2}\ln|x+1| + C$

$= \frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C$

Answer: $\frac{1}{2}\ln\left|\frac{x-1}{x+1}\right| + C$

Example 2: Repeated Linear Factor

Integrate $\int \frac{2x+1}{x^2(x-1)}\,dx$

Step 1: Set up partial fractions

Denominator has $x^2$ (repeated) and $(x-1)$ (distinct).

$\frac{2x+1}{x^2(x-1)} = \frac{A}{x} + \frac{B}{x^2} + \frac{C}{x-1}$

Step 2: Find A, B, C

Multiply by $x^2(x-1)$:

$2x+1 = Ax(x-1) + B(x-1) + Cx^2$

Let $x = 1$: $3 = C(1)$ → $C = 3$

Let $x = 0$: $1 = B(-1)$ → $B = -1$

Let $x = 2$: $5 = A(2)(1) + (-1)(1) + 3(4)$ $5 = 2A - 1 + 12$ $2A = -6$ → $A = -3$

Step 3: Integrate

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

$= -3\ln|x| + \frac{1}{x} + 3\ln|x-1| + C$

$= 3\ln\left|\frac{x-1}{x}\right| + \frac{1}{x} + C$

Answer: $3\ln\left|\frac{x-1}{x}\right| + \frac{1}{x} + C$

Example 3: Irreducible Quadratic

Integrate $\int \frac{x}{(x-1)(x^2+1)}\,dx$

Step 1: Set up partial fractions

$\frac{x}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$

Note: $x^2+1$ gets a linear numerator!

Step 2: Find A, B, C

Multiply by $(x-1)(x^2+1)$:

$x = A(x^2+1) + (Bx+C)(x-1)$

Let $x = 1$: $1 = A(2)$ → $A = \frac{1}{2}$

Expand and compare coefficients: $x = Ax^2 + A + Bx^2 - Bx + Cx - C$

Coefficient of $x^2$: $0 = A + B$ → $B = -\frac{1}{2}$

Coefficient of $x$: $1 = -B + C$ → $C = 1 - \frac{1}{2} = \frac{1}{2}$

Step 3: Integrate

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

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

For $\int \frac{-x/2}{x^2+1}\,dx$, use $u = x^2+1$:

$-\frac{1}{2}\int \frac{x}{x^2+1}\,dx = -\frac{1}{4}\ln(x^2+1)$

For $\int \frac{1/2}{x^2+1}\,dx$:

$\frac{1}{2}\arctan x$

Combine:

$= \frac{1}{2}\ln|x-1| - \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\arctan x + C$

Answer: $\frac{1}{2}\ln|x-1| - \frac{1}{4}\ln(x^2+1) + \frac{1}{2}\arctan x + C$

Improper Fractions: Long Division First

If degree of numerator ≥ degree of denominator, divide first!

Example: $\frac{x^3}{x^2-1}$

$\frac{x^3}{x^2-1} = x + \frac{x}{x^2-1}$

Then use partial fractions on $\frac{x}{x^2-1}$.

Method for Finding Constants

Method 1: Substitution (Cover-Up)

Substitute strategic values of $x$ to make terms disappear.

Works best for: distinct linear factors

Method 2: Comparing Coefficients

Expand and match coefficients of like powers.

Works for: all cases, especially quadratics

Method 3: Combination

Use substitution for easy ones, coefficients for the rest.

Most common approach!

⚠️ Common Mistakes

Mistake 1: Forgetting Long Division

If degree(numerator) ≥ degree(denominator), must divide first!

Mistake 2: Wrong Form for Quadratics

WRONG: $\frac{A}{x^2+1}$

RIGHT: $\frac{Ax+B}{x^2+1}$ (linear numerator!)

Mistake 3: Missing Powers in Repeated Factors

For $(x-1)^3$, you need: $\frac{A}{x-1} + \frac{B}{(x-1)^2} + \frac{C}{(x-1)^3}$

Don't skip the middle terms!

Mistake 4: Not Factoring Completely

Always factor the denominator completely first!

$x^2-1 = (x-1)(x+1)$, not just $x^2-1$

Summary of Decomposition Rules

📝 Practice Strategy

1. Check if fraction is proper (degree top < degree bottom)
2. If improper, do long division first
3. Factor denominator completely
4. Set up partial fraction form based on factors
5. Find constants using substitution and/or coefficients
6. Rewrite integral as sum of simpler integrals
7. Integrate each piece (use u-sub for quadratics if needed)
8. Combine and simplify

Step 2: Set up partial fractions.

$\frac{3x + 5}{(x+1)(x+2)} = \frac{A}{x+1} + \frac{B}{x+2}$

Multiply both sides by $(x+1)(x+2)$:

$3x + 5 = A(x+2) + B(x+1)$

Step 3: Solve for $A$ and $B$.

Method 1 (substitution):

• Let $x = -1$: $-3 + 5 = A(1)$, so $A = 2$
• Let $x = -2$: $-6 + 5 = B(-1)$, so $B = 1$

Step 4: Integrate.

$\int \frac{3x + 5}{x^2 + 3x + 2} \, dx = \int \left(\frac{2}{x+1} + \frac{1}{x+2}\right) dx$

$= 2\ln|x+1| + \ln|x+2| + C$

$= \ln|x+1|^2 + \ln|x+2| + C$

$= \ln|(x+1)^2(x+2)| + C$

Step 2: Find A, B, C

Multiply by $x(x-1)^2$:

$x^2+1 = A(x-1)^2 + Bx(x-1) + Cx$

Let $x = 0$: $1 = A(1)$ → $A = 1$

Let $x = 1$: $2 = C(1)$ → $C = 2$

Let $x = 2$: $5 = A(1) + B(2)(1) + C(2)$ $5 = 1 + 2B + 4$ $2B = 0$ → $B = 0$

Step 3: Integrate

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

$= \ln|x| + 2 \cdot \frac{(x-1)^{-1}}{-1} + C$

$= \ln|x| - \frac{2}{x-1} + C$

Answer: $\ln|x| - \frac{2}{x-1} + C$

Actually, let me redo this:

$x^3 + x = x(x^2+4) - 4x + x = x(x^2+4) - 3x$

$\frac{x^3+x}{x^2+4} = x - \frac{3x}{x^2+4}$

Step 3: Integrate

$\int \frac{x^3+x}{x^2+4}\,dx = \int x\,dx - 3\int \frac{x}{x^2+4}\,dx$

For the second integral, use $u = x^2+4$, $du = 2x\,dx$:

$\int \frac{x}{x^2+4}\,dx = \frac{1}{2}\ln(x^2+4)$

Step 4: Combine

$= \frac{x^2}{2} - 3 \cdot \frac{1}{2}\ln(x^2+4) + C$

$= \frac{x^2}{2} - \frac{3}{2}\ln(x^2+4) + C$

Answer: $\frac{x^2}{2} - \frac{3}{2}\ln(x^2+4) + C$