Before applying partial fractions or integrating a rational function, check if the degree of the numerator is greater than or equal to the degree of the denominator. If so, you must perform polynomial long division first to rewrite the integrand as a polynomial plus a proper rational function (where the numerator's degree is less than the denominator's degree).
🎯 In this section you will learn
- When to use polynomial long division before integrating
- How to perform polynomial long division step by step
- How to integrate the resulting polynomial and remainder
📌 Why Use Polynomial Long Division?
Consider $\int \frac{x^3}{x-1} dx$. The numerator (degree 3) has a higher degree than the denominator (degree 1). We cannot integrate this directly. But if we divide, we get:
$$ \frac{x^3}{x-1} = x^2 + x + 1 + \frac{1}{x-1} $$
Now we can integrate term by term!
💡 The Goal
Rewrite $\frac{P(x)}{Q(x)}$ as $\text{polynomial} + \frac{R(x)}{Q(x)}$ where degree of $R(x)$ < degree of $Q(x)$.
📋 Step-by-Step Polynomial Long Division
①
Divide the leading term
Divide the leading term of the numerator by the leading term of the denominator.
②
Multiply and subtract
Multiply the result by the denominator and subtract from the numerator.
③
Repeat
Repeat with the remainder until the remainder's degree is less than the denominator's degree.
Divide $\frac{x^3}{x-1}$.
①
Divide $x^3$ by $x$
$x^3 ÷ x = x^2$
②
Multiply and subtract
$x^2(x-1) = x^3 - x^2$
$x^3 - (x^3 - x^2) = x^2$
③
Bring down next term (none)
Remainder: $x^2$
④
Divide $x^2$ by $x$
$x^2 ÷ x = x$
⑤
Multiply and subtract
$x(x-1) = x^2 - x$
$x^2 - (x^2 - x) = x$
⑥
Divide $x$ by $x$
$x ÷ x = 1$
⑦
Multiply and subtract
$1(x-1) = x - 1$
$x - (x - 1) = 1$
⑧
Result
Quotient: $x^2 + x + 1$, Remainder: $1$
$\frac{x^3}{x-1} = x^2 + x + 1 + \frac{1}{x-1}$
Find $\displaystyle \int \frac{x^3}{x-1} dx$.
①
Perform long division
$\frac{x^3}{x-1} = x^2 + x + 1 + \frac{1}{x-1}$
②
Integrate term by term
$\int (x^2 + x + 1) dx + \int \frac{1}{x-1} dx$
③
Compute each integral
$\int x^2 dx = \frac{x^3}{3}$, $\int x dx = \frac{x^2}{2}$, $\int 1 dx = x$, $\int \frac{1}{x-1} dx = \ln|x-1|$
④
Combine
$\int \frac{x^3}{x-1} dx = \frac{x^3}{3} + \frac{x^2}{2} + x + \ln|x-1| + C$
Divide $\frac{2x^3 + 3x^2 - 4x + 1}{x^2 + 1}$.
①
Divide $2x^3$ by $x^2$
$2x^3 ÷ x^2 = 2x$
②
Multiply and subtract
$2x(x^2+1) = 2x^3 + 2x$
$(2x^3 + 3x^2 - 4x + 1) - (2x^3 + 2x) = 3x^2 - 6x + 1$
③
Divide $3x^2$ by $x^2$
$3x^2 ÷ x^2 = 3$
④
Multiply and subtract
$3(x^2+1) = 3x^2 + 3$
$(3x^2 - 6x + 1) - (3x^2 + 3) = -6x - 2$
⑤
Result
Quotient: $2x + 3$, Remainder: $-6x - 2$
$\frac{2x^3 + 3x^2 - 4x + 1}{x^2 + 1} = 2x + 3 + \frac{-6x - 2}{x^2 + 1}$
⚠️ Common Mistakes
- Forgetting to check degrees: Always check if the numerator's degree is ≥ denominator's degree before integrating.
- Missing terms with zero coefficients: When dividing, include placeholders for missing terms (e.g., $x^3 + 0x^2 + 0x - 1$).
- Stopping too early: Keep dividing until the remainder's degree is less than the denominator's degree.
🔍 Key Takeaways
- Use polynomial long division when degree(Numerator) ≥ degree(Denominator).
- The result is: polynomial + (proper rational function).
- After division, integrate the polynomial and the remainder separately.
- This technique is essential before partial fractions for improper rational functions.
← Back to Module Page