Partial Fractions for MATH 141
Exam Relevance for MATH 141
Partial fractions are a MATH 141 staple. Expect rational functions with linear and quadratic factors on both midterm and final.
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Understanding Partial Fractions
When you need to integrate a rational function like $\int \frac{3x+5}{x^2-x-2} \, dx$, there's no direct formula to apply. The denominator is too complicated.
Partial fractions is a technique that breaks apart a complex fraction into simpler pieces. Instead of one ugly fraction, you get a sum of easy fractions that you already know how to integrate!
The idea: $\frac{3x+5}{x^2-x-2}$ becomes $\frac{A}{x-2} + \frac{B}{x+1}$
Each piece integrates to a logarithm. Much easier!
When to Use Partial Fractions
Use partial fractions when:
- You have a rational function (polynomial ÷ polynomial)
- The degree of the numerator is less than the degree of the denominator
- The denominator can be factored
⚠️ If the numerator's degree ≥ denominator's degree, do polynomial long division first, then use partial fractions on the remainder.
The Cases
| Denominator Factor | Partial Fraction Form |
|---|---|
| Linear: $(ax + b)$ | $\frac{A}{ax+b}$ |
| Repeated linear: $(ax+b)^2$ | $\frac{A}{ax+b} + \frac{B}{(ax+b)^2}$ |
| Repeated linear: $(ax+b)^3$ | $\frac{A}{ax+b} + \frac{B}{(ax+b)^2} + \frac{C}{(ax+b)^3}$ |
| Irreducible quadratic: $(ax^2+bx+c)$ | $\frac{Ax+B}{ax^2+bx+c}$ |
Problem: Find $\int \frac{3x+5}{x^2-x-2} \, dx$
Step 1: Factor the denominator
$$x^2 - x - 2 = (x-2)(x+1)$$
Step 2: Set up partial fractions
$$\frac{3x+5}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$$
Step 3: Clear denominators
Multiply both sides by $(x-2)(x+1)$:
$$3x + 5 = A(x+1) + B(x-2)$$
Step 4: Solve for A and B
Method: Plug in convenient values
Let $x = 2$: $3(2) + 5 = A(3) + B(0)$
$$11 = 3A \implies A = \frac{11}{3}$$
Let $x = -1$: $3(-1) + 5 = A(0) + B(-3)$
$$2 = -3B \implies B = -\frac{2}{3}$$
Step 5: Integrate
$$\int \frac{3x+5}{x^2-x-2} \, dx$$
$$= \int \frac{11/3}{x-2} \, dx + \int \frac{-2/3}{x+1} \, dx$$
$$= \frac{11}{3}\ln|x-2| - \frac{2}{3}\ln|x+1| + C$$
$$= \boxed{\frac{11}{3}\ln|x-2| - \frac{2}{3}\ln|x+1| + C}$$
Problem: Find $\int \frac{5x+1}{(x+2)^2} \, dx$
Step 1: Set up partial fractions
For a repeated factor $(x+2)^2$, we need:
$$\frac{5x+1}{(x+2)^2} = \frac{A}{x+2} + \frac{B}{(x+2)^2}$$
Step 2: Clear denominators
$$5x + 1 = A(x+2) + B$$
Step 3: Solve for A and B
Let $x = -2$: $5(-2) + 1 = A(0) + B$
$$-9 = B$$
Expand and compare coefficients of $x$:
$$5x + 1 = Ax + 2A + B$$
Coefficient of $x$: $5 = A$
So $A = 5$, $B = -9$.
Step 4: Integrate
$$\int \frac{5x+1}{(x+2)^2} \, dx$$
$$= \int \frac{5}{x+2} \, dx + \int \frac{-9}{(x+2)^2} \, dx$$
$$= 5\ln|x+2| + \frac{9}{x+2} + C$$
$$= \boxed{5\ln|x+2| + \frac{9}{x+2} + C}$$
Problem: Find $\int \frac{2x}{(x-1)(x^2+1)} \, dx$
Step 1: Check if quadratic factors
$x^2 + 1$ doesn't factor over real numbers (irreducible).
Step 2: Set up partial fractions
$$\frac{2x}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{Bx+C}{x^2+1}$$
Note: Irreducible quadratics get $Bx + C$ on top, not just $B$.
Step 3: Clear denominators
$$2x = A(x^2+1) + (Bx+C)(x-1)$$
Step 4: Solve for A, B, C
Let $x = 1$: $2 = A(2) + 0$
$$A = 1$$
Expand the right side:
$$2x = Ax^2 + A + Bx^2 - Bx + Cx - C$$
$$2x = (A+B)x^2 + (-B+C)x + (A-C)$$
Compare coefficients:
- $x^2$: $0 = A + B = 1 + B \implies B = -1$
- $x^1$: $2 = -B + C = 1 + C \implies C = 1$
- $x^0$: $0 = A - C = 1 - 1$ ✓
Step 5: Integrate
$$\int \frac{2x}{(x-1)(x^2+1)} \, dx$$
$$= \int \frac{1}{x-1} \, dx + \int \frac{-x+1}{x^2+1} \, dx$$
Split the second integral:
$$= \int \frac{1}{x-1} \, dx - \int \frac{x}{x^2+1} \, dx + \int \frac{1}{x^2+1} \, dx$$
$$= \ln|x-1| - \frac{1}{2}\ln(x^2+1) + \arctan x + C$$
$$= \boxed{\ln|x-1| - \frac{1}{2}\ln(x^2+1) + \arctan x + C}$$
Problem: Find $\int \frac{x^2+2}{x-1} \, dx$
Step 1: Check degrees
Numerator degree (2) ≥ Denominator degree (1)
We must do long division first!
Step 2: Polynomial long division
$$\frac{x^2+2}{x-1} = x + 1 + \frac{3}{x-1}$$
(Check: $(x+1)(x-1) + 3 = x^2 - 1 + 3 = x^2 + 2$ ✓)
Step 3: Integrate
$$\int \frac{x^2+2}{x-1} \, dx = \int \left(x + 1 + \frac{3}{x-1}\right) dx$$
$$= \frac{x^2}{2} + x + 3\ln|x-1| + C$$
$$= \boxed{\frac{x^2}{2} + x + 3\ln|x-1| + C}$$
Finding Coefficients: Two Methods
Method 1: Strategic Values (Cover-Up)
Plug in values that make factors zero.
For $\frac{3x+5}{(x-2)(x+1)} = \frac{A}{x-2} + \frac{B}{x+1}$:
- Set $x = 2$ to find $A$ (kills the $B$ term)
- Set $x = -1$ to find $B$ (kills the $A$ term)
Method 2: Comparing Coefficients
Expand everything, then match coefficients of like powers.
When to use which:
- Strategic values: Quick for distinct linear factors
- Comparing coefficients: Necessary for repeated or quadratic factors
Common Mistakes and Misunderstandings
❌ Mistake: Forgetting repeated factor terms
Wrong setup: For $\frac{1}{(x+1)^2}$, writing just $\frac{A}{(x+1)^2}$
Why it's wrong: Repeated factors need multiple terms.
Correct: $\frac{A}{x+1} + \frac{B}{(x+1)^2}$
❌ Mistake: Using just a constant for quadratic factors
Wrong: $\frac{1}{(x-1)(x^2+1)} = \frac{A}{x-1} + \frac{B}{x^2+1}$
Why it's wrong: Irreducible quadratics need $Bx + C$, not just $B$.
Correct: $\frac{A}{x-1} + \frac{Bx+C}{x^2+1}$
❌ Mistake: Skipping long division
Wrong: Trying partial fractions on $\frac{x^3}{x^2-1}$ directly
Why it's wrong: Numerator degree (3) > denominator degree (2). Partial fractions requires the numerator to have smaller degree.
Correct: Do long division first, then partial fractions on the remainder.
Partial Fractions: Linear Factor
Each distinct linear factor in the denominator gets one term with an unknown constant A in the numerator.
Variables:
- $A$:
- unknown constant to solve for
- $ax+b$:
- the linear factor from the denominator
Partial Fractions: Repeated Linear Factor
A repeated linear factor (ax+b)^n needs n separate terms, with powers from 1 up to n.
Variables:
- $n$:
- the power of the repeated factor
- $A_1, A_2, ..., A_n$:
- unknown constants (one for each power)
Partial Fractions: Irreducible Quadratic
An irreducible quadratic (one that doesn't factor) gets Ax+B in the numerator, not just a constant.
Variables:
- $A, B$:
- unknown constants to solve for
- $ax^2+bx+c$:
- the quadratic that cannot be factored
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