AP® Calculus BC review sheet from Aim for Five (aimforfive.com/calc-bc/units/6/6-10)
Unit 6 · Topic 6.10
6.10 Integrating Functions Using Long Division and Completing the Square
Some rational functions don't fit any basic rule until you rearrange them. Long division handles fractions whose top has degree at least as big as the bottom's, and completing the square turns certain quadratic denominators into an arctangent form.
Key terms
- polynomial long division
- completing the square
- rational function
- arctangent
- degree
When to use long division
A rational function is a fraction of two polynomials. If the degree of the numerator is greater than or equal to the degree of the denominator, divide first. The result is a polynomial (easy to integrate) plus a remainder fraction whose top has a smaller degree than its bottom.
Example of the pattern: (x² + 3x + 5)/(x + 1) = x + 2 + 3/(x + 1), because (x + 1)(x + 2) = x² + 3x + 2, which leaves a remainder of 3. Now integrate each piece. The last one is a natural log: ∫ 3/(x + 1) dx = 3 ln|x + 1| + C.
A quick shortcut when the denominator is simple: add and subtract inside the numerator. For x/(x + 2), write (x + 2 − 2)/(x + 2) = 1 − 2/(x + 2).
When to complete the square
If the denominator is a quadratic with no real roots (its discriminant b² − 4ac is negative), you can't factor it. Complete the square to rewrite it as (x − h)² + a², then use
∫ 1/(a² + u²) du = (1/a) arctan(u/a) + C.
To complete the square for x² + bx + c, add and subtract (b/2)². For example, x² + 4x + 13 = (x² + 4x + 4) + 9 = (x + 2)² + 3².
A related form uses arcsine: ∫ 1/√(a² − u²) du = arcsin(u/a) + C, for a > 0. Completing the square can turn a root like √(−x² + 2x + 3) = √(4 − (x − 1)²) into this form.
Checklist for a rational integrand
- Is the top a constant multiple of the bottom's derivative? Then use u-substitution and get a log.
- Is the top's degree ≥ the bottom's? Divide first.
- Is the bottom an irreducible quadratic? Complete the square and look for arctan.
- Does the bottom factor into different linear factors? In BC, use partial fractions (6.12).
Why this topic matters
Both techniques are pure algebra followed by rules you already know (ln and arctan). They test whether you can spot that a messy integrand is a familiar one in disguise. That's a skill the course calls rearranging into equivalent forms, and it shows up again when you choose techniques in 6.14.
Worked examples
Try each one yourself first, then open the solution.
- Example 1
Long division first
Find ∫ (x² + 3x + 5)/(x + 1) dx.
Show the solutionHide the solution
- Step 1: Numerator degree 2 ≥ denominator degree 1, so divide.
- Step 2: x² + 3x + 5 = (x + 1)(x + 2) + 3, so the integrand is x + 2 + 3/(x + 1).
- Step 3: Integrate each term: x²/2 + 2x + 3 ln|x + 1|.
Answer: x²/2 + 2x + 3 ln|x + 1| + C
- Example 2
Completing the square → arctan
Find ∫ 1/(x² + 4x + 13) dx.
Show the solutionHide the solution
- Step 1: The discriminant is 16 − 52 < 0, so the denominator doesn't factor.
- Step 2: Complete the square: x² + 4x + 13 = (x + 2)² + 9.
- Step 3: Let u = x + 2, du = dx, and a = 3: ∫ 1/(u² + 9) du = (1/3) arctan(u/3) + C.
- Step 4: Check: d/dx [(1/3) arctan((x + 2)/3)] = (1/3) · (1/3)/(1 + (x + 2)²/9) = 1/(9 + (x + 2)²). That matches.
Answer: (1/3) arctan((x + 2)/3) + C
- Example 3
Exam level: division that leads to arctan
Evaluate ∫₀¹ x²/(x² + 1) dx.
Show the solutionHide the solution
- Step 1: Degrees are equal, so divide: x²/(x² + 1) = (x² + 1 − 1)/(x² + 1) = 1 − 1/(x² + 1).
- Step 2: Antiderivative: x − arctan x.
- Step 3: Evaluate: (1 − arctan 1) − (0 − arctan 0) = 1 − π/4.
- Step 4: Trap: x²/(x² + 1) is not ln(x² + 1) or anything like it. The top isn't a multiple of the bottom's derivative (2x).
Answer: 1 − π/4 ≈ 0.215
Common mistakes
- Trying u-substitution with u = x + 1 on (x² + 3x + 5)/(x + 1) and getting stuck. If the top's degree is at least the bottom's, divide.
- Forgetting the 1/a factor in (1/a) arctan(u/a).
- Completing the square incorrectly, for example writing x² + 4x + 13 = (x + 2)² + 13. Subtract the (b/2)² you added: 13 − 4 = 9.
- Writing ln of a quadratic just because the integrand is a fraction. The answer is a multiple of ln|bottom| only when the top is a multiple of the bottom's derivative.
On the exam
- These usually appear as no-calculator multiple-choice questions. The answer choices often include ln and arctan versions, so decide which pattern fits before computing.
- Check an answer by differentiating it, especially for arctan, where the constants are easy to slip on.
Connected topics
Videos
Check yourself
4 questions on 6.10 Integrating Functions Using Long Division and Completing the Square. Pick an answer to see if you got it, and why.
∫ (x² + 3x + 1)/(x + 1) dx =
∫ 1/(x² + 4x + 13) dx =
∫ x²/(x² + 1) dx =
∫₀¹ 1/(x² + 2x + 2) dx =
0 of 4 answered